ICSE Class 10 Mathematics Chapter 21 - Trigonometric Identities Selina Solutions

Exercise - 21(A)

1. Prove the following identity:

$\dfrac{{{\mathbf{sec}}\;{\mathbf{A}}\; - \;1}}{{{\mathbf{sec}}\;{\mathbf{A}}\; + \;1}} = \dfrac{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}$

Ans: Taking LHS -

${\text{LHS}}\,{\text{ = }}\,\dfrac{{{\text{sec A - 1}}}}{{{\text{sec A + 1}}}}$

$ = \dfrac{{\dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}\; - \;1}}{{\dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}\; + \;1}}$                     $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{cos}}\;{\mathbf{A}}\; = \dfrac{1}{{{\mathbf{sec}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{\dfrac{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}}}{{\dfrac{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}}}$ 

$ = \dfrac{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

2. Prove the following identity:

$\dfrac{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}} = \dfrac{{{\mathbf{cosec}}\;{\mathbf{A}}\; + \;1}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; - \;1}}$

Ans: Taking RHS -

${\mathbf{RHS}} = \dfrac{{{\mathbf{cosec}}\;{\mathbf{A}}\; + \;1}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; - \;1}}$

$ = \dfrac{{\dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}\; + \;1}}{{\dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}\; - \;1}}$                     $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{cosec}}\;{\mathbf{A}}\; = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{\dfrac{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}}}{{\dfrac{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}}}$ 

$ = \dfrac{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}} = {\mathbf{LHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

3. Prove the following identity:

$\dfrac{1}{{{\mathbf{tan}}\;{\mathbf{A}}\; + \;{\mathbf{cot}}\;\;{\mathbf{A}}}} = \;{\mathbf{cos}}\;{\mathbf{A}}\;{\mathbf{sin}}\;{\mathbf{A}}\;$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{1}{{{\mathbf{tan}}\;{\mathbf{A}}\; + \;{\mathbf{cot}}\;\;{\mathbf{A}}}}$

$ = \dfrac{1}{{\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}\; + \;\dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}}}$                   $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{tan}}\;{\mathbf{A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}},\;{\mathbf{cot}}\;{\mathbf{A}}\; = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{1}{{\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}\;{\mathbf{cos}}\;{\mathbf{A}}}}}}$ 

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\;{\mathbf{cos}}\;{\mathbf{A}}}}{{\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{1}}}$ 

$ = {\mathbf{sin}}\;{\mathbf{A}}\;{\mathbf{cos}}\;{\mathbf{A}}$                  $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = {\mathbf{cos}}\;{\mathbf{A}}\;{\mathbf{sin}}\;{\mathbf{A}}\; = {\mathbf{RHS}}$ 

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

4. Prove the following identity:

${\mathbf{tan}}\;{\mathbf{A}} - {\mathbf{cot}}\;{\mathbf{A}} = \dfrac{{1\; - \;2\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}\;{\mathbf{cos}}\;{\mathbf{A}}}}$

Ans: Taking LHS -

${\mathbf{LHS}} = {\mathbf{tan}}\;{\mathbf{A}} - {\mathbf{cot}}\;{\mathbf{A}}$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}} - \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}$               $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{tan}}\;{\mathbf{A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}},\;{\mathbf{cot}}\;{\mathbf{A}}\; = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}\;{\mathbf{sin}}\;{\mathbf{A}}}}$

$ = \dfrac{{1\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}\;{\mathbf{sin}}\;{\mathbf{A}}}}$             $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{{1\; - \;2\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}\;{\mathbf{cos}}\;{\mathbf{A}}}} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

5. Prove the following identity:

${\mathbf{si}}{{\mathbf{n}}^4}{\mathbf{A}} - {\mathbf{co}}{{\mathbf{s}}^4}{\mathbf{A}} = 2{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} - 1$

Ans: Taking LHS -

${\text{LHS}} = {\text{si}}{{\text{n}}^4}{\text{A}} - {\text{co}}{{\text{s}}^4}{\text{A}}$

$ = \left( {{\text{si}}{{\text{n}}^2}{\text{A}} - {\text{co}}{{\text{s}}^2}{\text{A}}} \right)\left( {{\text{si}}{{\text{n}}^2}{\text{A}} + {\text{co}}{{\text{s}}^2}{\text{A}}} \right)$         $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \left( {{\text{si}}{{\text{n}}^2}{\text{A}} - {\text{co}}{{\text{s}}^2}{\text{A}}} \right)$                        $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = {\text{si}}{{\text{n}}^2}{\text{A}} - \left( {1 - {\text{si}}{{\text{n}}^2}{\text{A}}} \right)$                 $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = 2{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} - 1 = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

6.  Prove the following identity:

${\left( {1 - {\mathbf{tan}}\;{\mathbf{A}}} \right)^2} + {\left( {1 + {\mathbf{tan}}\;{\mathbf{A}}} \right)^2} = 2{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = {\left( {1 - {\mathbf{tan}}\;{\mathbf{A}}} \right)^2} + {\left( {1 + {\mathbf{tan}}\;{\mathbf{A}}} \right)^2}$

$ = 1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} - 2{\mathbf{tan}}\;{\mathbf{A}} + 1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} + 2{\mathbf{tan}}\;{\mathbf{A}}$ $\left[ {{{\left( {{\mathbf{a}} \pm {\mathbf{b}}} \right)}^2} = {{\mathbf{a}}^2} + {{\mathbf{b}}^2} \pm 2{\mathbf{ab}}} \right]$

$ = 2 + 2{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}$

$ = 2\left( {1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}} \right)$

$ = 2{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}} = {\mathbf{RHS}}$                           $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} = {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}} \right]$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

7. Prove the following identity:

${\mathbf{cose}}{{\mathbf{c}}^4}{\mathbf{A}} - {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}} = {\mathbf{co}}{{\mathbf{t}}^4}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = {\mathbf{cose}}{{\mathbf{c}}^4}{\mathbf{A}} - {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}$

Taking ${\text{cose}}{{\text{c}}^2}{\text{A}}$ common,

$ = {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}\left( {{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}} - 1} \right)$

$ = {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}} \times {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}}$                $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;1 + {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} = {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}} \right]$

$ = (1 + {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}}) \times {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}}$        $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;1 + {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} = {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}} \right]$

$ = {\mathbf{co}}{{\mathbf{t}}^4}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

8.  Prove the following identity:

${\mathbf{sec}}\;{\mathbf{A}}\left( {1 - {\mathbf{sin}}\;{\mathbf{A}}} \right)\left( {{\mathbf{sec}}\;{\mathbf{A}} + {\mathbf{tan}}\;{\mathbf{A}}} \right) = 1$

Ans: Taking LHS -

${\mathbf{LHS}} = {\mathbf{sec}}\;{\mathbf{A}}\left( {1 - {\mathbf{sin}}\;{\mathbf{A}}} \right)\left( {{\mathbf{sec}}\;{\mathbf{A}} + {\mathbf{tan}}\;{\mathbf{A}}} \right)$

$ = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}\left( {1 - {\mathbf{sin}}\;{\mathbf{A}}} \right)\left( {\dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right)$ $\left[ {{\mathbf{We}}\;{\mathbf{know}}:\;{\mathbf{tan}}\;{\mathbf{A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}},\;{\mathbf{sec}}\;{\mathbf{A}}\; = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{\left( {1\; - \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}}{{{\mathbf{cos}}\;{\mathbf{A}}}}\left( {\dfrac{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right)$

$ = \dfrac{{\left( {{1^2}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}} \right)}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$                                       $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \dfrac{{\left( {1\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}} \right)}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$                                              $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = 1 = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

9. Prove the following identity:

${\mathbf{cosec}}\;{\mathbf{A}}\left( {1 + {\mathbf{cos}}\;{\mathbf{A}}} \right)\left( {{\mathbf{cosec}}\;{\mathbf{A}} - {\mathbf{cot}}\;{\mathbf{A}}} \right) = 1$

Ans: Taking LHS -

${\mathbf{LHS}} = {\mathbf{cosec}}\;{\mathbf{A}}\left( {1 + {\mathbf{cos}}\;{\mathbf{A}}} \right)\left( {{\mathbf{cosec}}\;{\mathbf{A}} - {\mathbf{cot}}\;{\mathbf{A}}} \right)$

$ = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}\left( {1 + {\mathbf{cos}}\;{\mathbf{A}}} \right)\left( {\dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}} - \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right)$               $\left[ {\;{\mathbf{cot}}\;{\mathbf{A}} = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}},\;{\mathbf{cosec}}\;{\mathbf{A}}\; = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{\left( {1\; + \;{\mathbf{cos}}\;{\mathbf{A}}} \right)}}{{{\mathbf{sin}}\;{\mathbf{A}}}}\left( {\dfrac{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right)$

$ = \dfrac{{\left( {{1^2}\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}} \right)}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$                                       $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \dfrac{{\left( {1\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}} \right)}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$                                              $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = 1 = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

10. Prove the following identity:

${\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}} + {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}} = {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}.{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{1}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}} + \dfrac{1}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$             $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{sec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}},\;{\mathbf{cosec}}\;{\mathbf{A}}\; = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; \times \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$

$ = \dfrac{1}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; \times \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$                           $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}.{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}} = {\mathbf{RHS}}$      $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{sec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}},\;{\mathbf{cosec}}\;{\mathbf{A}}\; = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

11. Prove the following identity:

$\dfrac{{\left( {1\; + \;{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}} \right)\;{\mathbf{cot}}\;{\mathbf{A}}}}{{{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}}} = {\mathbf{tan}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{\left( {1\; + \;{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}} \right)\;{\mathbf{cot}}\;{\mathbf{A}}}}{{{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}}}$

$ = \dfrac{{\left( {{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}} \right)\;{\mathbf{cot}}\;{\mathbf{A}}}}{{{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}}}$                                  $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} = {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}} \right]$

$ = \dfrac{{\dfrac{1}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}} \times \;\dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}}}{{\dfrac{1}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}}}$                         $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{sec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}},\;{\mathbf{cosec}}\;{\mathbf{A}}\; = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{\dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}} \times \;\dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}}}{{\dfrac{1}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}}}$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}$                                      $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{tan}}\;{\mathbf{A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$

$ = {\mathbf{tan}}\;{\mathbf{A}} = {\mathbf{RHS}}$ 

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

12.  Prove the following identity:

${\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} - {\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} = {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}.{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}} - \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{1}$             $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{tan}}\;{\mathbf{A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; \times \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\;\left( {1 - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}} \right)}}{{\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$                          

$ = {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}.\left( {{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}} \right) = {\mathbf{RHS}}\;$  $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1,{\mathbf{tan}}\;{\mathbf{A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

13. Prove the following identity:

${\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} - {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}.{\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}}$

Ans: Taking LHS -

 ${\mathbf{LHS}} = \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}} - \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{1}$             $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{cot}}\;{\mathbf{A}} = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; \times \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\;\left( {1 - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}} \right)}}{{\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$                          

$ = {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}}.\left( {{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}} \right) = {\mathbf{RHS}}\;$  $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1,{\mathbf{cot}}\;{\mathbf{A}} = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

14. Prove the following identity:

$\left( {{\mathbf{cosec}}\;{\mathbf{A}} + {\mathbf{sin}}\;{\mathbf{A}}} \right)\left( {{\mathbf{cosec}}\;{\mathbf{A}} - {\mathbf{sin}}\;{\mathbf{A}}} \right) = {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \left( {{\mathbf{cosec}}\;{\mathbf{A}} + {\mathbf{sin}}\;{\mathbf{A}}} \right)\left( {{\mathbf{cosec}}\;{\mathbf{A}} - {\mathbf{sin}}\;{\mathbf{A}}} \right)$

$ = \left( {{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}} - {\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}} \right)$                      $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1\;\& \;1 + {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} = {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}} \right]$

$ = 1 + {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} - \left( {1 - {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}} \right)$ 

$ = {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

15. Prove the following identity:

$\left( {{\mathbf{sec}}\;{\mathbf{A}} - {\mathbf{cos}}\;{\mathbf{A}}} \right)\left( {{\mathbf{sec}}\;{\mathbf{A}} + {\mathbf{cos}}\;{\mathbf{A}}} \right) = {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}$

Ans: Taking LHS -

 ${\mathbf{LHS}} = \left( {{\mathbf{sec}}\;{\mathbf{A}} - {\mathbf{cos}}\;{\mathbf{A}}} \right)\left( {{\mathbf{sec}}\;{\mathbf{A}} + {\mathbf{cos}}\;{\mathbf{A}}} \right)$

$ = \left( {{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}} - {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}} \right)$                      $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1\;\& \;1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} = {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}} \right]$

$ = 1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} - \left( {1 - {\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}} \right)$ 

$ = {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

16. Prove the following identity:

${\left( {{\mathbf{cos}}\;{\mathbf{A}} + {\mathbf{sin}}\;{\mathbf{A}}} \right)^2} + {\left( {{\mathbf{cos}}\;{\mathbf{A}} - {\mathbf{sin}}\;{\mathbf{A}}} \right)^2} = 2$

Ans: Taking LHS -

${\mathbf{LHS}} = {\left( {{\mathbf{cos}}\;{\mathbf{A}} + {\mathbf{sin}}\;{\mathbf{A}}} \right)^2}{\left( {{\mathbf{cos}}\;{\mathbf{A}} - {\mathbf{sin}}\;{\mathbf{A}}} \right)^2}$

$ = {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} + {\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + 2{\mathbf{cos}}\;{\mathbf{A}}.{\mathbf{sin}}\;{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} + {\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} - 2\;{\mathbf{cos}}\;{\mathbf{A}}.{\mathbf{sin}}\;{\mathbf{A}}$

$\left[ {{{\left( {{\mathbf{a}} \pm {\mathbf{b}}} \right)}^2} = {{\mathbf{a}}^2} + {{\mathbf{b}}^2} \pm 2{\mathbf{ab}}} \right]$

$ = 2\left( {{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} + {\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}} \right)$                $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = 2 = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

17. Prove the following identity:

$\left( {{\mathbf{cosec}}\;{\mathbf{A}} - {\mathbf{sin}}\;{\mathbf{A}}} \right)\left( {{\mathbf{sec}}\;{\mathbf{A}} - {\mathbf{cos}}\;{\mathbf{A}}} \right)\left( {{\mathbf{tan}}\;{\mathbf{A}} + {\mathbf{cot}}\;{\mathbf{A}}} \right) = 1$

Ans: Taking LHS -

${\mathbf{LHS}} = \left( {{\mathbf{cosec}}\;{\mathbf{A}} - {\mathbf{sin}}\;{\mathbf{A}}} \right)\left( {{\mathbf{sec}}\;{\mathbf{A}} - {\mathbf{cos}}\;{\mathbf{A}}} \right)\left( {{\mathbf{tan}}\;{\mathbf{A}} + {\mathbf{cot}}\;{\mathbf{A}}} \right)$

$\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{cot}}\;{\mathbf{A}} = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}},\;{\mathbf{tan}}\;{\mathbf{A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}},\;{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}},\;{\mathbf{sec}}\; = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$

$ = \left( {\dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}} - {\mathbf{sin}}\;{\mathbf{A}}} \right)\left( {\dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}} - {\mathbf{cos}}\;{\mathbf{A}}} \right)\left( {\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right)$ 

$ = \left( {\dfrac{{1\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right)\left( {\dfrac{{1\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right)\left( {\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}\;.\;{\mathbf{sin}}\;{\mathbf{A}}}}} \right)$

$ = \left( {\dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right)\left( {\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right)\left( {\dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}\;.\;{\mathbf{sin}}\;{\mathbf{A}}}}} \right)$           $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = 1 = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

18. Prove the following identity:

$\dfrac{1}{{{\mathbf{sec}}\;{\mathbf{A}}\; + \;{\mathbf{tan}}\;{\mathbf{A}}}} = {\mathbf{sec}}\;{\mathbf{A}} - {\mathbf{tan}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{1}{{{\mathbf{sec}}\;{\mathbf{A}}\; + \;{\mathbf{tan}}\;{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}\; - \;{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{sec}}\;{\mathbf{A}}\; + \;{\mathbf{tan}}\;{\mathbf{A}}}}$                  $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} = {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}} \right]$

$ = \dfrac{{\left( {{\mathbf{sec}}\;{\mathbf{A}}\; - \;{\mathbf{tan}}\;{\mathbf{A}}} \right)\left( {{\mathbf{sec}}\;{\mathbf{A}}\; + \;{\mathbf{tan}}\;{\mathbf{A}}} \right)}}{{{\mathbf{sec}}\;{\mathbf{A}}\; + \;{\mathbf{tan}}\;{\mathbf{A}}}}$                $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = {\mathbf{sec}}\;{\mathbf{A}} - {\mathbf{tan}}\;{\mathbf{A}} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

19.  Prove the following identity:

${\mathbf{cosec}}\;{\mathbf{A}} + {\mathbf{cot}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{cosec}}\;{\mathbf{A}}\; - \;{\mathbf{cot}}\;{\mathbf{A}}}}$

Ans: Taking RHS -

${\mathbf{RHS}} = \dfrac{1}{{{\mathbf{cosec}}\;{\mathbf{A}}\; - \;{\mathbf{cot}}\;{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}\; - \;{\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}}}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; - \;{\mathbf{cot}}\;{\mathbf{A}}}}$                  $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;1 + {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} = {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}} \right]$

$ = \dfrac{{\left( {{\mathbf{cosec}}\;{\mathbf{A}}\; - \;{\mathbf{cot}}\;{\mathbf{A}}} \right)\left( {{\mathbf{cosec}}\;{\mathbf{A}}\; + \;{\mathbf{cot}}\;{\mathbf{A}}} \right)}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; - \;{\mathbf{cot}}\;{\mathbf{A}}}}$                $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = {\mathbf{cosec}}\;{\mathbf{A}} + {\mathbf{cot}}\;{\mathbf{A}} = {\mathbf{LHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

20. Prove the following identity:

$\dfrac{{{\mathbf{sec}}\;{\mathbf{A}}\; - \;{\mathbf{tan}}\;{\mathbf{A}}}}{{{\mathbf{sec}}\;{\mathbf{A}}\; + \;{\mathbf{tan}}\;{\mathbf{A}}}} = 1 - 2\;{\mathbf{sec}}\;{\mathbf{A}}.{\mathbf{tan}}\;{\mathbf{A}} + 2\;{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{{\mathbf{sec}}\;{\mathbf{A}}\; - \;{\mathbf{tan}}\;{\mathbf{A}}}}{{{\mathbf{sec}}\;{\mathbf{A}}\; + \;{\mathbf{tan}}\;{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{sec}}\;{\mathbf{A}}\; - \;{\mathbf{tan}}\;{\mathbf{A}}}}{{{\mathbf{sec}}\;{\mathbf{A}}\; + \;{\mathbf{tan}}\;{\mathbf{A}}}} \times \dfrac{{{\mathbf{sec}}\;{\mathbf{A}}\; - \;{\mathbf{tan}}\;{\mathbf{A}}}}{{{\mathbf{sec}}\;{\mathbf{A}}\; - \;{\mathbf{tan}}\;{\mathbf{A}}}}$

$ = \dfrac{{{{\left( {{\mathbf{sec}}\;{\mathbf{A}}\; - \;{\mathbf{tan}}\;{\mathbf{A}}} \right)}^2}}}{{{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}\; - \;{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}}}$

$\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}} - {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} = 1\;\& \;\;{{\left( {{\mathbf{a}} + {\mathbf{b}}} \right)}^2} = {{\mathbf{a}}^2} + {{\mathbf{b}}^2} + 2{\mathbf{ab}}} \right]$

$ = {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}} + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} - 2\;{\mathbf{sec}}\;{\mathbf{A}}.\;{\mathbf{tan}}\;{\mathbf{A}}$   

$\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}} = 1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}} \right]$

$ = 1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} - 2\;{\mathbf{sec}}\;{\mathbf{A}}.\;{\mathbf{tan}}\;{\mathbf{A}}$ 

$ = 1 - 2\;{\mathbf{sec}}\;{\mathbf{A}}.\;{\mathbf{tan}}\;{\mathbf{A}} + 2\;{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

21. Prove the following identity:

${\left( {{\mathbf{sin}}\;{\mathbf{A}} + {\mathbf{cosec}}\;{\mathbf{A}}} \right)^2} + {\left( {{\mathbf{cos}}\;{\mathbf{A}} + {\mathbf{sec}}\;{\mathbf{A}}} \right)^2} = 7 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = {\left( {{\mathbf{sin}}\;{\mathbf{A}} + {\mathbf{cosec}}\;{\mathbf{A}}} \right)^2}{\left( {{\mathbf{cos}}\;{\mathbf{A}} + {\mathbf{sec}}\;{\mathbf{A}}} \right)^2}$

$ = {\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}} + 2{\mathbf{cosec}}\;{\mathbf{A}}.{\mathbf{sin}}\;{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} + {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}} + 2\;{\mathbf{cos}}\;{\mathbf{A}}.{\mathbf{sec}}\;{\mathbf{A}}$

$\left[ {{{\left( {{\mathbf{a}} + {\mathbf{b}}} \right)}^2} = {{\mathbf{a}}^2} + {{\mathbf{b}}^2} + 2{\mathbf{ab}}} \right]$

$ = \left( {{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} + {\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}} \right) + \left( {{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}} + {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}} \right) + 2 + 2\;$ $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{cosec}}\;{\mathbf{A}}.{\mathbf{sin}}\;{\mathbf{A}} = 1\;\& \;{\mathbf{cos}}\;{\mathbf{A}}.{\mathbf{sec}}\;{\mathbf{A}} = 1} \right]$

$ = 2 + 2 + 1 + \left( {1 + {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}}} \right) + \left( {1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}} \right)$  

$\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}} = 1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}\;\& \;1 + {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} = {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}} \right]$

$ = 7 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

22. Prove the following identity:

${\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}.{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}} = {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} + 2$

Ans: Taking LHS -

${\mathbf{LHS}} = {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}.{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}$

$ = \dfrac{1}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}} \times \dfrac{1}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\;.\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$                                $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\;.\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}} + \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\;.\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$

$ = \dfrac{1}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}} + \dfrac{1}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$

$ = {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}} + {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}$               $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}},\;{\mathbf{sec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$

$ = 1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} + 1 + {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}}\;$       $\left[ {{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}} = 1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}\;\& \;1 + {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} = {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}} \right]$

$ = {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} + 2 = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

23. Prove the following identity:

$\dfrac{1}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}} + \;\dfrac{1}{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}} = 2\;{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{1}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}} + \;\dfrac{1}{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}$

$ = \dfrac{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}\; + \;1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}{{\left( {1\; - \;{\mathbf{cos}}\;{\mathbf{A}}} \right)\;\left( {1\; + \;{\mathbf{cos}}\;{\mathbf{A}}} \right)}}$

$ = \dfrac{2}{{{1^2}\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$                              $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \dfrac{2}{{1\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$ 

$ = \dfrac{2}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$                                    $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = 2\;{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}} = {\mathbf{RHS}}$              $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$ 

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

24. Prove the following identity:

$\dfrac{1}{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}} + \;\dfrac{1}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}} = 2\;{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{1}{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}} + \;\dfrac{1}{{1 + \;{\mathbf{sin}}\;{\mathbf{A}}}}$

$ = \dfrac{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}\; + \;1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}{{\left( {1\; - \;{\mathbf{sin}}\;{\mathbf{A}}} \right)\;\left( {1\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}}$

$ = \dfrac{2}{{{1^2}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$                              $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \dfrac{2}{{1\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$ 

$ = \dfrac{2}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$                                    $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = 2\;{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}} = {\mathbf{RHS}}$              $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{sec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$ 

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

25. Prove the following identity:

$\dfrac{{{\mathbf{cosec}}\;{\mathbf{A}}}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; - \;1}} + \;\dfrac{{{\mathbf{cosec}}\;{\mathbf{A}}}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; + \;1}} = 2\;{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{{\mathbf{cosec}}\;{\mathbf{A}}}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; - \;1}} + \;\dfrac{{{\mathbf{cosec}}\;{\mathbf{A}}}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; + \;1}}$

$ = {\mathbf{cosec}}\;{\mathbf{A}} \times \left[ {\dfrac{{1\; + \;{\mathbf{cosec}}\;{\mathbf{A}}\; + \;{\mathbf{cosec}}\;{\mathbf{A}}\; - \;1}}{{\left( {{\mathbf{cosec}}\; - \;1} \right)\;\left( {{\mathbf{cosec}}\; + \;1} \right)}}} \right]$

$ = \dfrac{{2\;{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}\;}}{{{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}\; - \;{1^2}}}$                              $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \dfrac{{2\;\;{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}}}{{{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}\; - \;1}}$ 

$ = \dfrac{{2\;\;{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}}}{{{\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}}}}$                                   $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;1 + {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} = {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}} \right]$

$ = 2 \times \dfrac{{\dfrac{1}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}}}{{\dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}}}$

$ = 2 \times \dfrac{1}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$

$ = 2\;{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}} = {\mathbf{RHS}}$              $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{sec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$ 

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

26. Prove the following identity:

$\dfrac{{{\mathbf{sec}}\;{\mathbf{A}}}}{{{\mathbf{sec}}\;{\mathbf{A}}\; + \;1}} + \;\dfrac{{{\mathbf{sec}}\;{\mathbf{A}}}}{{{\mathbf{sec}}\;{\mathbf{A}}\; - \;1}} = 2\;{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{{\mathbf{sec}}\;{\mathbf{A}}}}{{{\mathbf{sec}}\;{\mathbf{A}}\; + \;1}} + \;\dfrac{{{\mathbf{sec}}\;{\mathbf{A}}}}{{{\mathbf{sec}}\;{\mathbf{A}}\; - \;1}}$

$ = {\mathbf{sec}}\;{\mathbf{A}} \times \left[ {\dfrac{{{\mathbf{sec}}\;{\mathbf{A}}\; - 1\; + \;{\mathbf{sec}}\;{\mathbf{A}}\; + \;1}}{{\left( {{\mathbf{sec}}\; + \;1} \right)\;\left( {{\mathbf{sec}}\; - \;1} \right)}}} \right]$

$ = \dfrac{{2\;{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}\;}}{{{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}\; - \;{1^2}}}$                              $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \dfrac{{2\;\;{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}}}{{{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}\; - \;{1^{}}}}$ 

$ = \dfrac{{2\;\;{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}}}{{{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}}}$                                   $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} = {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}} \right]$

$ = 2 \times \dfrac{{\dfrac{1}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}}}{{\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}}}$

$ = 2 \times \dfrac{1}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$

$ = 2\;{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}} = {\mathbf{RHS}}$              $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$ 

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

27.  Prove the following identity:

$\dfrac{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}} = \;\dfrac{{{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{{\left( {{\mathbf{sec}}\;{\mathbf{A}}\; - \;1} \right)}^2}}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}$

Divide N & D by cos A,

$ = \dfrac{{\dfrac{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}}}{{\dfrac{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}}}$                          

$ = \dfrac{{\dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}\; + \;1}}{{\dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}\; - \;{1^{}}}}$ 

$ = \dfrac{{{\mathbf{sec}}\;{\mathbf{A}}\; + \;1}}{{{\mathbf{sec}}\;{\mathbf{A}}\; - \;1}}$                            $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{sec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{{\mathbf{sec}}\;{\mathbf{A}}\; + \;1}}{{{\mathbf{sec}}\;{\mathbf{A}}\; - \;1}} \times \dfrac{{{\mathbf{sec}}\;{\mathbf{A}}\; - \;1}}{{{\mathbf{sec}}\;{\mathbf{A}}\; - \;1}}$

$ = \dfrac{{{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}\; - \;{1^2}}}{{{{\left( {{\mathbf{sec}}\;{\mathbf{A}}\; - \;1} \right)}^2}}}$                                   $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \dfrac{{{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}\; - \;1}}{{{{\left( {{\mathbf{sec}}\;{\mathbf{A}}\; - \;1} \right)}^2}}}$

$ = \dfrac{{{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{{\left( {{\mathbf{sec}}\;{\mathbf{A}}\; - \;1} \right)}^2}}} = {\mathbf{RHS}}$                       $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} = {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}} \right]$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

28. Prove the following identity:

$\dfrac{{{\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}}}}{{{{\left( {{\mathbf{cosec}}\;{\mathbf{A}}\; + \;1} \right)}^2}}} = \;\dfrac{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}$

Ans: Taking RHS -

${\mathbf{RHS}} = \;\dfrac{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}$

Divide N & D by sin A,

$ = \dfrac{{\dfrac{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}}}{{\dfrac{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}}}$                          

$ = \dfrac{{\dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}\; - \;1}}{{\dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}\; + \;{1^{}}}}$ 

$ = \dfrac{{{\mathbf{cosec}}\;{\mathbf{A}}\; - \;1}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; + \;1}}$                            $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{{\mathbf{cosec}}\;{\mathbf{A}}\; - \;1}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; + \;1}} \times \dfrac{{{\mathbf{cosec}}\;{\mathbf{A}}\; + \;1}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; + \;1}}$

$ = \dfrac{{{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}\; - \;{1^2}}}{{{{\left( {{\mathbf{cosec}}\;{\mathbf{A}}\; + \;1} \right)}^2}}}$                                   $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \dfrac{{{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}\; - \;1}}{{{{\left( {{\mathbf{cosec}}\;{\mathbf{A}}\; + \;1} \right)}^2}}}$

$ = \dfrac{{{\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}}}}{{{{\left( {{\mathbf{cosec}}\;{\mathbf{A}}\; + \;1} \right)}^2}}} = {\mathbf{LHS}}$                       $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;1 + {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} = {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}} \right]$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

29. Prove the following identity:

$\dfrac{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}} = \;2\;{\mathbf{sec}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}$

$ = \dfrac{{{{\left( {1\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}^2} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}\left( {1\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}}$

$ = \dfrac{{1\; + \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;2\;{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}\left( {1\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}}$                    $\left[ {{{\left( {{\mathbf{a}} + {\mathbf{b}}} \right)}^2} = {{\mathbf{a}}^2} + {{\mathbf{b}}^2} + 2{\mathbf{ab}}} \right]$

$ = \dfrac{{1\; + \;1\; + \;2\;{\mathbf{sin}}\;{\mathbf{A}}\;}}{{{\mathbf{cos}}\;{\mathbf{A}}\left( {1\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}}$                                 $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{{2\left( {1\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}}{{{\mathbf{cos}}\;{\mathbf{A}}\left( {1\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}}$

$ = \dfrac{2}{{{\mathbf{cos}}\;{\mathbf{A}}}}$

$ = 2\;{\mathbf{secA}} = {\mathbf{RHS}}$                            $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{sec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

30. Prove the following identity:

$\dfrac{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}} = {\left( {{\mathbf{sec}}\;{\mathbf{A}} - {\mathbf{tan}}\;{\mathbf{A}}} \right)^2}$

Ans: Taking LHS -

${\mathbf{LHS}} = \;\dfrac{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}$

Divide N & D by cos A,

$ = \dfrac{{\dfrac{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}}}{{\dfrac{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}}}$                          

$ = \dfrac{{\dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}\; - \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}\;}}{{\dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}\; + \;\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}}}$ 

$ = \dfrac{{{\mathbf{sec}}\;{\mathbf{A}}\; - \;{\mathbf{tan}}\;{\mathbf{A}}}}{{{\mathbf{sec}}\;{\mathbf{A}}\; + \;{\mathbf{tan}}\;{\mathbf{A}}}}$                            $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{sec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}},\;{\mathbf{tan}}\;{\mathbf{A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{{\mathbf{sec}}\;{\mathbf{A}}\; - \;{\mathbf{tan}}\;{\mathbf{A}}}}{{{\mathbf{sec}}\;{\mathbf{A}}\; + \;{\mathbf{tan}}\;{\mathbf{A}}}} \times \dfrac{{{\mathbf{sec}}\;{\mathbf{A}}\; - \;{\mathbf{tan}}\;{\mathbf{A}}}}{{{\mathbf{sec}}\;{\mathbf{A}}\; - \;{\mathbf{tan}}\;{\mathbf{A}}}}$

$ = \dfrac{{{{\left( {{\mathbf{sec}}\;{\mathbf{A}} - {\mathbf{tan}}\;{\mathbf{A}}} \right)}^2}}}{{{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}\; - \;{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}}}$                                   $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \dfrac{{{{\left( {{\mathbf{sec}}\;{\mathbf{A}} - {\mathbf{tan}}\;{\mathbf{A}}} \right)}^2}}}{1}$                       $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} = {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}} \right]$

$ = {\left( {{\mathbf{sec}}\;{\mathbf{A}} - {\mathbf{tan}}\;{\mathbf{A}}} \right)^2} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

31.  Prove the following identity:

${\left( {{\mathbf{cot}}\;{\mathbf{A}} - {\mathbf{cosec}}\;{\mathbf{A}}} \right)^2} = \dfrac{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}$

Ans: Taking RHS -

Divide N & D by sin A,

$ = \dfrac{{\dfrac{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}}}{{\dfrac{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}}}$                          

$ = \dfrac{{\dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}\; - \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}\;}}{{\dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}\; + \;\dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}}}$ 

$ = \dfrac{{{\mathbf{cosec}}\;{\mathbf{A}}\; - \;{\mathbf{cot}}\;{\mathbf{A}}}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; + \;{\mathbf{cot}}\;{\mathbf{A}}}}$                        $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}},\;{\mathbf{cot}}\;{\mathbf{A}} = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{{\mathbf{cosec}}\;{\mathbf{A}}\; - \;{\mathbf{cot}}\;{\mathbf{A}}}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; + \;{\mathbf{cot}}\;{\mathbf{A}}}} \times \dfrac{{{\mathbf{cosec}}\;{\mathbf{A}}\; - \;{\mathbf{cot}}\;{\mathbf{A}}}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; - \;{\mathbf{cot}}\;{\mathbf{A}}}}$

$ = \dfrac{{{{\left( {{\mathbf{cosec}}\;{\mathbf{A}}\; - \;{\mathbf{cot}}\;{\mathbf{A}}} \right)}^2}}}{{{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}\; - \;{\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}}}}$                                   $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \dfrac{{{{\left[ { - \left( {{\mathbf{cot}}\;{\mathbf{A}}\; - \;{\mathbf{cosec}}\;{\mathbf{A}}} \right)} \right]}^2}}}{1}$                       $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;1 + {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} = {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}} \right]$

$ = {\left( {{\mathbf{cot}}\;{\mathbf{A}}\; - \;{\mathbf{cosec}}\;{\mathbf{A}}} \right)^2} = {\mathbf{LHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

32. Prove the following identity:

$\dfrac{{{\mathbf{cosec}}\;{\mathbf{A}}\; - \;1}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; + \;1}} = {\left( {\dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}} \right)^2}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{\dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}\; - \;1}}{{\dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}\; + \;1}}$              $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{\dfrac{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}}}{{\dfrac{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}}}$

$ = \dfrac{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}$

$ = \dfrac{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}} \times \dfrac{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}$

$ = \dfrac{{{1^2} - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{{\left( {1\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}^2}}}$                                     $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \dfrac{{1\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{{\left( {1\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}^2}}}$

$ = \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{{\left( {1\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}^2}}}$                                         $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = {\left( {\dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}} \right)^2} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

33. Prove the following identity:

${\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} - {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{B}} = \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\;.\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}}}$

Ans: Taking LHS -

${\mathbf{LHS}} = {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} - {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{B}}$

$ = \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}} - \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}}}$                                       $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{tan}}\;{\mathbf{A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\;.\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}\;.\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}.{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}}}$

$ = \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\;\left( {1\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}} \right)\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}\;\left( {1\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}} \right)}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}.{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}}}$               $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\;\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\;.\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}\; + \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\;.\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}.{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}}}$

$ = \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\;.\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}}} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

34. Prove the following identity:

$\dfrac{{{\mathbf{sin}}\;{\mathbf{\theta }}\; - \;2\;{\mathbf{si}}{{\mathbf{n}}^3}{\mathbf{\theta }}}}{{2\;{\mathbf{co}}{{\mathbf{s}}^3}{\mathbf{\theta }}\; - \;{\mathbf{cos}}\;{\mathbf{\theta }}}} = {\mathbf{tan}}\;{\mathbf{\theta }}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{\theta }}\; - \;2\;{\mathbf{si}}{{\mathbf{n}}^3}{\mathbf{\theta }}}}{{2\;{\mathbf{co}}{{\mathbf{s}}^3}{\mathbf{\theta }}\; - \;{\mathbf{cos}}\;{\mathbf{\theta }}}}$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{\theta }}}}{{{\mathbf{cos}}\;{\mathbf{\theta }}}}\left[ {\dfrac{{1\; - \;2\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{\theta }}}}{{2\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{\theta }}\; - \;1}}} \right]$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{\theta }}}}{{{\mathbf{cos}}\;{\mathbf{\theta }}}}\left[ {\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{\theta }}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{\theta }}\; - \;2\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{\theta }}}}{{2\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{\theta }}\; - \;({\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{\theta }}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{\theta }})}}} \right]$                 $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = {\mathbf{tan}}\;{\mathbf{\theta }}\left[ {\dfrac{{\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{\theta }}\; - \;\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{\theta }}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{\theta }}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{\theta }}}}} \right]$

$ = {\mathbf{tan}}\;{\mathbf{\theta }} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

35. Prove the following identity:

$\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}} = {\mathbf{cosec}}\;{\mathbf{A}} - {\mathbf{cot}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}} \times \dfrac{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\left( {1\; - \;{\mathbf{cos}}\;{\mathbf{A}}} \right)}}{{{1^2}\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$                                 $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}} - {\mathbf{sin}}\;{\mathbf{A}}\;.\;{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}} - \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\;.\;{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}} - \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\;.\;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$                        $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}} - \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}$

$ = {\mathbf{cosec}}\;{\mathbf{A}} - {\mathbf{cot}}\;{\mathbf{A}} = {\mathbf{RHS}}$    $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}},{\mathbf{cot}}\;{\mathbf{A}} = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

36. Prove the following identity:

$\dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}} = {\mathbf{sec}}\;{\mathbf{A}} + {\mathbf{tan}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}} \times \dfrac{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}\left( {1\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}}{{{1^2}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$                                 $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}\;.\;{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}} + \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\;.\;{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}} + \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\;.\;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$                        $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}$

$ = {\mathbf{sec}}\;{\mathbf{A}} + {\mathbf{tan}}\;{\mathbf{A}} = {\mathbf{RHS}}$    $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{sec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}},{\mathbf{tan}}\;{\mathbf{A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

37. Prove the following identity:

$\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\;.\;{\mathbf{tan}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}} = 1 + {\mathbf{sec}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\;.\;{\mathbf{tan}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\;.\;\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}}}{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}$                                 $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{tan}}\;{\mathbf{A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{\;\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}}}{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}$

$ = \dfrac{{\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}\left( {1\; - \;{\mathbf{cos}}\;{\mathbf{A}}} \right)}}$

$ = \dfrac{{\;1\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}\left( {1\; - \;{\mathbf{cos}}\;{\mathbf{A}}} \right)}}$                           $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{{\;\left( {1\; - \;{\mathbf{cos}}\;{\mathbf{A}}} \right)\left( {1\; + \;{\mathbf{cos}}\;{\mathbf{A}}} \right)}}{{{\mathbf{cos}}\;{\mathbf{A}}\left( {1\; - \;{\mathbf{cos}}\;{\mathbf{A}}} \right)}}$

$ = \dfrac{{\;\left( {1\; + \;{\mathbf{cos}}\;{\mathbf{A}}} \right)}}{{{\mathbf{cos}}\;{\mathbf{A}}}}$

$ = \dfrac{{\;1}}{{{\mathbf{cos}}\;{\mathbf{A}}}} + \dfrac{{\;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}$                          $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{sec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$

$ = 1 + {\mathbf{sec}}\;{\mathbf{A}} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

38. Prove the following identity:

$\left( {1 + {\mathbf{cot}}\;{\mathbf{A}} - {\mathbf{cosec}}\;{\mathbf{A}}} \right)\left( {1 + {\mathbf{tan}}\;{\mathbf{A}} + {\mathbf{sec}}\;{\mathbf{A}}} \right) = 2$

Ans: Taking LHS -

${\text{LHS}} = \left( {1 + {\mathbf{cot}}\;{\mathbf{A}} - {\mathbf{cosec}}\;{\mathbf{A}}} \right)\left( {1 + {\mathbf{tan}}\;{\mathbf{A}} + {\mathbf{sec}}\;{\mathbf{A}}} \right)$

$ = \left( {1 + \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}} - \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right)\left( {1 + \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}} + \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right)$ 

$\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{sec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}},{\mathbf{tan}}\;{\mathbf{A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}},\;{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}},{\mathbf{cot}} = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = \left( {\dfrac{{{\text{sin A}} + {\text{cos A}} - {\text{}}1}}{{{\text{sin A}}}}} \right)\left( {\dfrac{{{\text{cos A}} + {\text{sin A}} + {\text{}}1}}{{{\text{cos A}}}}} \right)$

$ = \dfrac{{{\text{cos A}}\left( {{\text{sin A}} + {\text{cos A}} - {\text{}}1} \right){\text{}} + {\text{sin A}}\left( {{\text{sin A}} + {\text{cos A}} - {\text{}}1} \right){\text{}} + {\text{sin A}} + {\text{cos A}} - {\text{}}1}}{{{\text{sin A}}.{\text{cos A}}}}$

$ = \dfrac{{{\cos A} {\sin A} + \;{{\cos }^2}A\ - \cos A + \;{{\sin }^2}A\; + \;\sin \;A\;\cos \;A\; - \;\sin \;A\; + \;\sin \;A\; + \;\cos \;A\; - \;1}}{{\sin \;A\;.\;\cos \;A}}$

$ = \dfrac{{{\text{cos A.sin A}} - {\text{cos A}} + {\text{}}1{\text{}} + {\text{sin A.cos A}} - {\text{sin A}} + {\text{sin A}} + {\text{cos A}} - {\text{}}1}}{{{\text{sin A}}.{\text{cos A}}}}$

 $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{{2{\text{cos A}}.{\text{sin A}}}}{{{\text{sin A}}.{\text{cos A}}}}$ 

$ = 2 = {\text{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

39. Prove the following identity:

$\sqrt {\dfrac{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}}  = {\mathbf{sec}}\;{\mathbf{A}} + {\mathbf{tan}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \sqrt {\dfrac{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}} $

$ = \sqrt {\dfrac{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}} \times \dfrac{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}} $

$ = \sqrt {\dfrac{{{{\left( {1\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}^2}}}{{{1^2}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}} $                       $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \sqrt {\dfrac{{{{\left( {1\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}^2}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}} \;\;$                     $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}$

$ = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}$                      $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{sec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}},{\mathbf{tan}}\;{\mathbf{A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$

$ = {\mathbf{sec}}\;{\mathbf{A}} + {\mathbf{tan}}\;{\mathbf{A}} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

40. Prove the following identity:

$\sqrt {\dfrac{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}}  = {\mathbf{cosec}}\;{\mathbf{A}} - {\mathbf{cot}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \sqrt {\dfrac{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}} $

$ = \sqrt {\dfrac{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}} \times \dfrac{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}} $

$ = \sqrt {\dfrac{{{{\left( {1\; - \;{\mathbf{cos}}\;{\mathbf{A}}} \right)}^2}}}{{{1^2}\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}} $                       $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \sqrt {\dfrac{{{{\left( {1\; - \;{\mathbf{cos}}\;{\mathbf{A}}} \right)}^2}}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}} \;\;$                     $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}$

$ = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}} - \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}$                    $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}},{\mathbf{cot}}\;{\mathbf{A}} = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = {\mathbf{cosec}}\;{\mathbf{A}} - {\mathbf{cot}}\;{\mathbf{A}} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

41. Prove the following identity:

$\sqrt {\dfrac{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}}  = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{1 + {\mathbf{cos}}\;{\mathbf{A}}}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \sqrt {\dfrac{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}} $

$ = \sqrt {\dfrac{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}} \times \dfrac{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}} $

$ = \sqrt {\dfrac{{{1^2}\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{{\left( {1\; + \;{\mathbf{cos}}\;{\mathbf{A}}} \right)}^2}}}} $                       $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \sqrt {\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{{\left( {1\; + \;{\mathbf{cos}}\;{\mathbf{A}}} \right)}^2}}}} \;\;$                     $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

43. Prove the following identity:

$1 - \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}} = {\mathbf{sin}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = 1 - \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}$

$ = 1 - \dfrac{{{1^2}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}$                                    $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = 1 - \left[ {\dfrac{{\left( {1\; - \;{\mathbf{sin}}\;{\mathbf{A}}} \right)\left( {1\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}} \right]\;$                   $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = 1 - \left[ {1 - {\mathbf{sin}}\;{\mathbf{A}}} \right]$

$ = {\mathbf{sin}}\;{\mathbf{A}} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

44.  Prove the following identity:

$\dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{A}}}} + \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{cos}}\;{\mathbf{A}}}} = \dfrac{{2\;{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; - \;2\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{A}}}} + \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}{{\left( {{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{A}}} \right)\left( {{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{cos}}\;{\mathbf{A}}} \right)}}$                                   

$ = \dfrac{{2{\mathbf{sin}}\;{\mathbf{A}}}}{{\left( {{\mathbf{sin}}{\;^2}{\mathbf{A}}\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}} \right)}}$                      $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \dfrac{{2{\mathbf{sin}}\;{\mathbf{A}}}}{{\left( {1\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}} \right)}}$                  $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{{2\;{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; - \;2\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

45. Prove the following identity:

$\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{cos}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{A}}}} = \dfrac{2}{{2\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;1}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{cos}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}$

$ = \dfrac{{{{\left( {{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{A}}} \right)}^2}\; + \;{{\left( {{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{cos}}\;{\mathbf{A}}} \right)}^2}}}{{\left( {{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{A}}} \right)\left( {{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{cos}}\;{\mathbf{A}}} \right)}}$                                   

$ = \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; + \;2\;{\mathbf{sin}}\;{\mathbf{A}}.\;{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; - \;\;2\;{\mathbf{sin}}\;{\mathbf{A}}.\;{\mathbf{cos}}\;{\mathbf{A}}}}{{\left( {{\mathbf{sin}}{\;^2}{\mathbf{A}}\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}} \right)}}$  $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \dfrac{{2\left( {{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}} \right)}}{{\left( {{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;(1\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}} \right)}}$                                        $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{2}{{2\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;1}} = {\mathbf{RHS}}$                                     $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

46. Prove the following identity:

$\dfrac{{{\mathbf{cot}}\;{\mathbf{A}}\; + \;{\mathbf{cosec}}\;{\mathbf{A}}\; - \;1}}{{{\mathbf{cot}}\;{\mathbf{A}}\; - \;{\mathbf{cosec}}\;{\mathbf{A}}\; + \;1}} = \dfrac{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{{\mathbf{cot}}\;{\mathbf{A}}\; + \;{\mathbf{cosec}}\;{\mathbf{A}}\; - \;1}}{{{\mathbf{cot}}\;{\mathbf{A}}\; - \;{\mathbf{cosec}}\;{\mathbf{A}}\; + \;1}}$

$ = \left[ {\dfrac{{{\mathbf{cot}}\;{\mathbf{A}}\; + \;{\mathbf{cosec}}\;{\mathbf{A}}\; - \;\left( {{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}\; - \;{\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}}} \right)\;}}{{{\mathbf{cot}}\;{\mathbf{A}}\; - \;{\mathbf{cosec}}\;{\mathbf{A}}\; + \;1}}} \right]$             $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;1 + {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} = {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}} \right]$

$ = \left[ {\dfrac{{{\mathbf{cot}}\;{\mathbf{A}}\; + \;{\mathbf{cosec}}\;{\mathbf{A}}\; - \;\left( {{\mathbf{cosec}}\;{\mathbf{A}}\; - \;{\mathbf{cotA}}} \right)\left( {{\mathbf{cot}}\;{\mathbf{A}}\; + \;{\mathbf{cosec}}\;{\mathbf{A}}} \right)\;}}{{{\mathbf{cot}}\;{\mathbf{A}}\; - \;{\mathbf{cosec}}\;{\mathbf{A}}\; + \;1}}} \right]$           $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \left[ {\dfrac{{\left( {{\mathbf{cot}}\;{\mathbf{A}}\; + \;{\mathbf{cosec}}\;{\mathbf{A}}} \right)\left( {1 - {\mathbf{cosec}}\;{\mathbf{A}}\; + \;{\mathbf{cotA}}} \right)\;}}{{{\mathbf{cot}}\;{\mathbf{A}}\; - \;{\mathbf{cosec}}\;{\mathbf{A}}\; + \;1}}} \right]$  

$ = {\mathbf{cot}}\;{\mathbf{A}} + {\mathbf{cosec}}\;{\mathbf{A}}$

$ = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}} + \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}$                         $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}},{\mathbf{cot}}\;{\mathbf{A}} = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

47. Prove the following identity:

$\dfrac{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; - \;{\mathbf{cot}}\;{\mathbf{A}}}} - \dfrac{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; + \;{\mathbf{cot}}\;{\mathbf{A}}}} = 2\left( {1 + {\mathbf{cot}}\;{\mathbf{A}}} \right)$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; - \;{\mathbf{cot}}\;{\mathbf{A}}}} - \dfrac{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; + \;{\mathbf{cot}}\;{\mathbf{A}}}}$

$ = \dfrac{{\left( {1\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)\left( {{\mathbf{cosec}}\;{\mathbf{A}}\; + \;{\mathbf{cot}}\;{\mathbf{A}}} \right)\; - \;\left( {1\; - \;{\mathbf{sin}}\;{\mathbf{A}}} \right)\left( {{\mathbf{cosec}}\;{\mathbf{A}}\; - \;{\mathbf{cot}}\;{\mathbf{A}}} \right)}}{{\left( {{\mathbf{cosec}}\;{\mathbf{A}}\; - \;{\mathbf{cot}}\;{\mathbf{A}}} \right)\left( {{\mathbf{cosec}}\;{\mathbf{A}}\; + \;{\mathbf{cot}}\;{\mathbf{A}}} \right)}}$

$ = \dfrac{{{\mathbf{cosec}}\;{\mathbf{A}}\; + \;{\mathbf{cot}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}\;{\mathbf{cosec}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}\;{\mathbf{cot}}\;{\mathbf{A}}\; - \;{\mathbf{cosec}}\;{\mathbf{A}}\; + \;{\mathbf{cot}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}\;{\mathbf{cosec}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{A}}\;{\mathbf{cot}}\;{\mathbf{A}}}}{{\left( {{\mathbf{cosec}}\;{\mathbf{A}}\; - \;{\mathbf{cot}}\;{\mathbf{A}}} \right)\left( {{\mathbf{cosec}}\;{\mathbf{A}}\; + \;{\mathbf{cot}}\;{\mathbf{A}}} \right)}}$$ = \dfrac{{2\;\left( {{\mathbf{cot}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}\;{\mathbf{cosec}}\;{\mathbf{A}}} \right)}}{{\left( {{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}\; - \;{\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}}} \right)}}$                                            $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \dfrac{{2\;\left( {{\mathbf{cot}}\;{\mathbf{A}}\; + \;1} \right)}}{1}$          $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;1 + {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} = {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}\;\& \;{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = 2\left( {1 + {\mathbf{cot}}\;{\mathbf{A}}} \right) = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

48. Prove the following identity:

$\dfrac{{{\mathbf{cos}}\;{\mathbf{\theta }}\;.\;{\mathbf{cot}}\;{\mathbf{\theta }}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{\theta }}}} = {\mathbf{cosec}}\;{\mathbf{\theta }} - 1$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{{\mathbf{cos}}\;{\mathbf{\theta }}\;.\;{\mathbf{cot}}\;{\mathbf{\theta }}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{\theta }}}}$

$ = \dfrac{{{\mathbf{cos}}\;{\mathbf{\theta }}\;.\;\dfrac{{{\mathbf{cos}}\;{\mathbf{\theta }}}}{{{\mathbf{sin}}\;{\mathbf{\theta }}}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{\theta }}}}$                                                  $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{cot}}\;{\mathbf{A}} = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{\dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{\theta }}}}{{{\mathbf{sin}}\;{\mathbf{\theta }}}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{\theta }}}}$

$ = \dfrac{{\dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{\theta }}}}{{{\mathbf{sin}}\;{\mathbf{\theta }}}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{\theta }}}} \times \dfrac{{1\; - \;{\mathbf{sin}}\;{\mathbf{\theta }}}}{{1\; - \;{\mathbf{sin}}\;{\mathbf{\theta }}}}$

$ = \dfrac{{\dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{\theta }}}}{{{\mathbf{sin}}\;{\mathbf{\theta }}}}\; \times \;\left( {1\; - \;{\mathbf{sin}}\;{\mathbf{\theta }}} \right)}}{{{1^2}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{\theta }}}}$                                            $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \dfrac{{\dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{\theta }}}}{{{\mathbf{sin}}\;{\mathbf{\theta }}}}\; \times \;\left( {1\; - \;{\mathbf{sin}}\;{\mathbf{\theta }}} \right)}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{\theta }}}}$                                           $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{{1\; - \;{\mathbf{sin}}\;{\mathbf{\theta }}}}{{{\mathbf{sin}}\;{\mathbf{\theta }}}}$        

$ = {\mathbf{cosec}}\;{\mathbf{\theta }} - 1 = {\mathbf{RHS}}$                              $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

49. Prove the following identity:

$\dfrac{{{\mathbf{sin}}\;{\mathbf{\theta }}\;.\;{\mathbf{tan}}\;{\mathbf{\theta }}}}{{1\; - \;{\mathbf{cos}}\;{\mathbf{\theta }}}} = {\mathbf{sec}}\;{\mathbf{\theta }} + 1$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{\theta }}\;.\;{\mathbf{tan}}\;{\mathbf{\theta }}}}{{1\; - \;{\mathbf{cos}}\;{\mathbf{\theta }}}}$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{\theta }}\;.\;\dfrac{{{\mathbf{sin}}\;{\mathbf{\theta }}}}{{{\mathbf{cos}}\;{\mathbf{\theta }}}}}}{{1\; - \;{\mathbf{cos}}\;{\mathbf{\theta }}}}$                                                  $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{tan}}\;{\mathbf{A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{\theta }}}}{{{\mathbf{cos}}\;{\mathbf{\theta }}}}}}{{1\; - \;{\mathbf{cos}}\;{\mathbf{\theta }}}}$

$ = \dfrac{{\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{\theta }}}}{{{\mathbf{cos}}\;{\mathbf{\theta }}}}}}{{1\; - \;{\mathbf{cos}}\;{\mathbf{\theta }}}} \times \dfrac{{1\; + \;{\mathbf{cos}}\;{\mathbf{\theta }}}}{{1\; + \;{\mathbf{cos}}\;{\mathbf{\theta }}}}$

$ = \dfrac{{\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{\theta }}}}{{{\mathbf{cos}}\;{\mathbf{\theta }}}}\; \times \;\left( {1\; + \;{\mathbf{cos}}\;{\mathbf{\theta }}} \right)}}{{{1^2}\; - \;{\mathbf{co}}{{\mathbf{s}}^2}\;{\mathbf{\theta }}}}$                                            $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \dfrac{{\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{\theta }}}}{{{\mathbf{cos}}\;{\mathbf{\theta }}}}\; \times \;\left( {1\; + \;{\mathbf{cos}}\;{\mathbf{\theta }}} \right)}}{{{\mathbf{si}}{{\mathbf{n}}^2}\;{\mathbf{\theta }}}}$                                         $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{{1\; + \;{\mathbf{cos}}\;{\mathbf{\theta }}}}{{{\mathbf{cos}}\;{\mathbf{\theta }}}}$        

$ = {\mathbf{sec}}\;{\mathbf{\theta }} + 1 = {\mathbf{RHS}}$                              $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{sec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

Exercise - 21(B)

1. Prove that:

i) $\dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{tan}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{cot}}\;{\mathbf{A}}}} = {\mathbf{sin}}\;{\mathbf{A}} + {\mathbf{cos}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{tan}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{cot}}\;{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; - \;\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}}} + \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; - \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}}}$                 $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{tan}}\;{\mathbf{A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}},\;{\mathbf{cot}}\;{\mathbf{A}} = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{\dfrac{{{\mathbf{cos}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}}} + \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}}}$

$ = \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{A}}}} - \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}$

$ = \dfrac{{\left( {{\mathbf{cos}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{A}}} \right)\left( {{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}}{{{\mathbf{cos}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}$                                       $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = {\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

ii) $\dfrac{{{\mathbf{co}}{{\mathbf{s}}^3}{\mathbf{A}}\; + \;{\mathbf{si}}{{\mathbf{n}}^3}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{co}}{{\mathbf{s}}^3}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^3}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{A}}}} = 2$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{{\mathbf{co}}{{\mathbf{s}}^3}{\mathbf{A}}\; + \;{\mathbf{si}}{{\mathbf{n}}^3}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{co}}{{\mathbf{s}}^3}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^3}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}$

$ = \dfrac{{\left( {{\mathbf{co}}{{\mathbf{s}}^3}{\mathbf{A}}\; + \;{\mathbf{si}}{{\mathbf{n}}^3}{\mathbf{A}}} \right)\left( {{\mathbf{cos}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{A}}} \right)\; + \;\left( {{\mathbf{co}}{{\mathbf{s}}^3}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^3}{\mathbf{A}}} \right)\left( {{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}}{{\left( {{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)\left( {{\mathbf{cos}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}}$

$ = \dfrac{{\left( {{\mathbf{co}}{{\mathbf{s}}^3}{\mathbf{A}}\; + \;{\mathbf{si}}{{\mathbf{n}}^3}{\mathbf{A}}} \right)\left( {{\mathbf{cos}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{A}}} \right)\; + \;\left( {{\mathbf{co}}{{\mathbf{s}}^3}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^3}{\mathbf{A}}} \right)\left( {{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$      $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \dfrac{{\left( {{\mathbf{co}}{{\mathbf{s}}^4}{\mathbf{A}}\; + \;{\mathbf{si}}{{\mathbf{n}}^3}{\mathbf{A}}.{\mathbf{cos}}\;{\mathbf{A}}\; - \;{\mathbf{co}}{{\mathbf{s}}^3}{\mathbf{A}}.\;{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^4}{\mathbf{A}}} \right)\; + \;\left( {{\mathbf{co}}{{\mathbf{s}}^4}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^3}{\mathbf{A}}\;.\;{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{co}}{{\mathbf{s}}^3}{\mathbf{A}}\;.\;{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^4}{\mathbf{A}}} \right)}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$$ = \dfrac{{2\;\left( {{\mathbf{co}}{{\mathbf{s}}^4}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^4}{\mathbf{A}}} \right)}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$

$ = \dfrac{{2\;\left( {{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}} \right)\left( {{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; + \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}} \right)}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$                        $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = 2\left( {{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; + \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}} \right)$                         $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = 2 = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

iii) $\dfrac{{{\mathbf{tan}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{cot}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{cot}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{tan}}\;{\mathbf{A}}}} = {\mathbf{sec}}\;{\mathbf{A}}.{\mathbf{cosec}}\;{\mathbf{A}} + 1$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{{\mathbf{tan}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{cot}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{cot}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{tan}}\;{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{tan}}\;{\mathbf{A}}}}{{1\; - \;\dfrac{1}{{{\mathbf{tan}}\;{\mathbf{A}}}}}} + \dfrac{{\dfrac{1}{{{\mathbf{tan}}\;{\mathbf{A}}}}}}{{1\; - \;{\mathbf{tan}}\;{\mathbf{A}}}}$                                     $\left[ {\;{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{cot}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{tan}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{tan}}\;{\mathbf{A}}\; - \;1}} + \dfrac{1}{{{\mathbf{tan}}\;{\mathbf{A}}\left( {1\; - \;{\mathbf{tan}}\;{\mathbf{A}}} \right)}}$

$ = \dfrac{{{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{tan}}\;{\mathbf{A}}\; - \;1}} - \dfrac{1}{{{\mathbf{tan}}\;{\mathbf{A}}\left( {{\mathbf{tan}}\;{\mathbf{A}}\; - \;1} \right)}}$

$ = \dfrac{{{\mathbf{ta}}{{\mathbf{n}}^3}{\mathbf{A}}\; - \;1}}{{{\mathbf{tan}}\;{\mathbf{A}}\left( {{\mathbf{tan}}\;{\mathbf{A}}\; - \;1} \right)}}$

$ = \dfrac{{\left( {{\mathbf{tan}}\;{\mathbf{A}}\; - \;1} \right)\left( {{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;1\; + \;{\mathbf{tan}}\;{\mathbf{A}}} \right)}}{{{\mathbf{tan}}\;{\mathbf{A}}\left( {{\mathbf{tan}}\;{\mathbf{A}}\; - \;1} \right)}}$                        $\left[ {{{\mathbf{a}}^3} - {{\mathbf{b}}^3} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{{\mathbf{a}}^2} + {{\mathbf{b}}^2} + {\mathbf{ab}}} \right)} \right]$

$ = \dfrac{{{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;1\; + \;{\mathbf{tan}}\;{\mathbf{A}}}}{{{\mathbf{tan}}\;{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}\;\; + \;{\mathbf{tan}}\;{\mathbf{A}}}}{{{\mathbf{tan}}\;{\mathbf{A}}}}$                                      $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} = {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}} \right]$

$ = \dfrac{{{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}}}{{{\mathbf{tan}}\;{\mathbf{A}}}} + 1$

$=\dfrac{{\dfrac{1}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}}}{{\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}}} + 1$                              $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{sec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}},\;{\mathbf{tan}}\;{\mathbf{A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}.\;{\mathbf{cos}}\;{\mathbf{A}}}} + 1$

$ = {\mathbf{cosec}}\;{\mathbf{A}}\;.\;{\mathbf{sec}}\;{\mathbf{A}} + 1 = {\mathbf{RHS}}\;$                     $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

iv) ${\left( {{\mathbf{tan}}\;{\mathbf{A}} + \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right)^2} + {\left( {{\mathbf{tan}}\;{\mathbf{A}} - \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right)^2} = 2\left( {\dfrac{{1\; + \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{1\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}} \right)$

Ans: Taking LHS -

${\mathbf{LHS}} = {\left( {{\mathbf{tan}}\;{\mathbf{A}} + \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right)^2} + {\left( {{\mathbf{tan}}\;{\mathbf{A}} - \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right)^2}$

$ = {\left( {\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}} + \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right)^2} + {\left( {\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}} - \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right)^2}$              $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{tan}}\;{\mathbf{A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$

$ = {\left( {\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\; + \;1}}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right)^2} + {\left( {\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\; - \;1}}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right)^2}$

$ = \dfrac{{{{\left( {{\mathbf{sin}}\;{\mathbf{A}}\; + \;1} \right)}^2}\; + \;{{\left( {{\mathbf{sin}}\;{\mathbf{A}}\; - \;1} \right)}^2}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;1\; + \;2\;{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;1\; - \;2\;{\mathbf{sin}}\;{\mathbf{A}}\;}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$                         $\left[ {{{\left( {{\mathbf{a}} \pm {\mathbf{b}}} \right)}^2} = {{\mathbf{a}}^2} + {{\mathbf{b}}^2} \pm 2{\mathbf{ab}}} \right]$

$ = \dfrac{{2\;\left( {\;1\; + \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\;} \right)}}{{1\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$                                              $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = 2\left( {\dfrac{{1\; + \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{1\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}} \right) = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

v) $2\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^4}{\mathbf{A}} = 1 + {\mathbf{si}}{{\mathbf{n}}^4}{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = 2\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^4}{\mathbf{A}}$

$ = 2\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\left( {1 - {\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}} \right)^2}$                         $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = 2\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + 1 + {\mathbf{si}}{{\mathbf{n}}^4}{\mathbf{A}} - 2\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\;$                        $\left[ {{{\left( {{\mathbf{a}} - {\mathbf{b}}} \right)}^2} = {{\mathbf{a}}^2} + {{\mathbf{b}}^2} - 2{\mathbf{ab}}} \right]$

$ = 1 + {\mathbf{si}}{{\mathbf{n}}^4}{\mathbf{A}} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

vi) $\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{B}}}}{{{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{B}}}} + \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}\; - \;{\mathbf{cos}}\;{\mathbf{B}}}}{{{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{B}}}} = 0$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{B}}}}{{{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{B}}}} + \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}\; - \;{\mathbf{cos}}\;{\mathbf{B}}}}{{{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{B}}}}$

$ = \dfrac{{\left( {{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{B}}} \right)\left( {{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{B}}} \right)\; + \;\left( {{\mathbf{cos}}\;{\mathbf{A}}\; - \;{\mathbf{cos}}\;{\mathbf{B}}} \right)\left( {{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{B}}} \right)}}{{\left( {{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{B}}} \right)\left( {{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{B}}} \right)}}$

$ = \dfrac{{\left( {{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}} \right)\; + \;\left( {{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}} \right)}}{{\left( {{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{B}}} \right)\left( {{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{B}}} \right)}}$                                    $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \dfrac{{\left( {{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}} \right)\; - \;\left( {{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}} \right)}}{{\left( {{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{B}}} \right)\left( {{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{B}}} \right)}}$         

$ = \dfrac{{1\; - \;1}}{{\left( {{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{B}}} \right)\left( {{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{B}}} \right)}}$                            $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{0}{{\left( {{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{B}}} \right)\left( {{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{B}}} \right)}}$

$ = 0 = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

vii) $\left( {{\mathbf{cosec}}\;{\mathbf{A}} - {\mathbf{sin}}\;{\mathbf{A}}} \right)\left( {{\mathbf{sec}}\;{\mathbf{A}} - {\mathbf{cos}}\;{\mathbf{A}}} \right) = \dfrac{1}{{{\mathbf{tan}}\;{\mathbf{A}}\; + \;{\mathbf{cot}}\;{\mathbf{A}}}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \left( {{\mathbf{cosec}}\;{\mathbf{A}} - {\mathbf{sin}}\;{\mathbf{A}}} \right)\left( {{\mathbf{sec}}\;{\mathbf{A}} - {\mathbf{cos}}\;{\mathbf{A}}} \right)$

$ = \left( {\dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}} - {\mathbf{sin}}\;{\mathbf{A}}} \right)\left( {\dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}} - {\mathbf{cos}}\;{\mathbf{A}}} \right)$              $\left[ {\;{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}},\;{\mathbf{sec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$

$ = \left( {\dfrac{{1\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right)\left( {\dfrac{{1\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right)$

$ = \left( {\dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right)\left( {\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right)$

$ = {\mathbf{sin}}\;{\mathbf{A}}\;.\;{\mathbf{cos}}\;{\mathbf{A}}$

Taking RHS - 

${\text{RHS}} = \dfrac{1}{{{\mathbf{tan}}\;{\mathbf{A}}\; + \;{\mathbf{cot}}\;{\mathbf{A}}}}$

$ = \dfrac{1}{{\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}\; + \;\dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}}}$                       $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{cot}}\;{\mathbf{A}} = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}},\;{\mathbf{tan}}\;{\mathbf{A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{1}{{\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}\; \times \;{\mathbf{sin}}\;{\mathbf{A}}}}}}$

$ = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}\; \times \;{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$                                       

$ = {\mathbf{sin}}\;{\mathbf{A}}\;.\;{\mathbf{cos}}\;{\mathbf{A}}$                                       $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

viii) ${\left( {1 + {\mathbf{tan}}\;{\mathbf{A}}.{\mathbf{tan}}\;{\mathbf{B}}} \right)^2} + {\left( {{\mathbf{tan}}\;{\mathbf{A}} - {\mathbf{tan}}\;{\mathbf{B}}} \right)^2} = {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}\;.\;{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{B}}$

Ans: Taking LHS -

${\mathbf{LHS}} = {\left( {1 + {\mathbf{tan}}\;{\mathbf{A}}.{\mathbf{tan}}\;{\mathbf{B}}} \right)^2} + {\left( {{\mathbf{tan}}\;{\mathbf{A}} - {\mathbf{tan}}\;{\mathbf{B}}} \right)^2}$

$ = 1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}.{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{B}} + 2\;{\mathbf{tan}}\;{\mathbf{A}}.{\mathbf{tan}}\;{\mathbf{B}} + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{B}} - 2\;{\mathbf{tan}}\;{\mathbf{A}}.{\mathbf{tan}}\;{\mathbf{B}}$

$\left[ {{{\left( {{\mathbf{a}} \pm {\mathbf{b}}} \right)}^2} = {{\mathbf{a}}^2} + {{\mathbf{b}}^2} \pm 2{\mathbf{ab}}} \right]$

$ = 1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}.{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{B}} + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{B}}$

$ = {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}} + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{B}}\left( {{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} + 1} \right)$                 $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} = {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}} \right]$

$ = {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}} + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{B}}\left( {{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}} \right)$                          $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} = {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}} \right]$

$ = {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}\left( {1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{B}}} \right)$

$ = {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}\;.\;{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{B}} = {\mathbf{RHS}}$                            $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} = {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}} \right]$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

ix) $\dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}\; - \;1}} + \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}\; + \;1}} = {\mathbf{cosec}}\;{\mathbf{A}} + {\mathbf{sec}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}\; - \;1}} + \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}\; + \;1}}$

$ = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}\; - \;1\; + \;{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}\; + \;1}}{{\left( {{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}\; - \;1} \right)\left( {{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}\; + \;1} \right)}}$

$ = \dfrac{{2\left( {{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}}{{{{\left( {{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}^2}\; - \;{1^2}}}$                                                     $\left[ {{{\mathbf{a}}^2} - {{\mathbf{b}}^2} = \left( {{\mathbf{a}} - {\mathbf{b}}} \right)\left( {{\mathbf{a}} + {\mathbf{b}}} \right)} \right]$

$ = \dfrac{{2\left( {{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; + \;2\;{\mathbf{sin}}\;{\mathbf{A}}.\;{\mathbf{cos}}\;{\mathbf{A}}\; - \;1}}$                                     $\left[ {{{\left( {{\mathbf{a}} + {\mathbf{b}}} \right)}^2} = {{\mathbf{a}}^2} + {{\mathbf{b}}^2} + 2{\mathbf{ab}}} \right]$

$ = \dfrac{{2\left( {{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}}{{1\; + \;2\;{\mathbf{sin}}\;{\mathbf{A}}.\;{\mathbf{cos}}\;{\mathbf{A}}\; - \;1}}$

$ = \dfrac{{\left( {{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}}{{{\mathbf{sin}}\;{\mathbf{A}}.\;{\mathbf{cos}}\;{\mathbf{A}}}}$

$ = \dfrac{{\left( {{\mathbf{cos}}\;{\mathbf{A}}} \right)}}{{{\mathbf{sin}}\;{\mathbf{A}}.\;{\mathbf{cos}}\;{\mathbf{A}}}} + \dfrac{{\left( {{\mathbf{sin}}\;{\mathbf{A}}} \right)}}{{{\mathbf{sin}}\;{\mathbf{A}}.\;{\mathbf{cos}}\;{\mathbf{A}}}}$

$ = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}} + \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}$

$ = {\mathbf{cosec}}\;{\mathbf{A}} + {\mathbf{sec}}\;{\mathbf{A}} = {\mathbf{RHS}}$     $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}},\;{\mathbf{sec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

2.If ${\mathbf{x}}.{\mathbf{cos}}\;{\mathbf{A}} + {\mathbf{y}}.{\mathbf{sin}}\;{\mathbf{A}} = {\mathbf{m}}$ and ${\mathbf{x}}.{\mathbf{sin}}\;{\mathbf{A}} - {\mathbf{y}}.{\mathbf{cos}}\;{\mathbf{A}} = {\mathbf{n}}$, then prove that: ${{\mathbf{x}}^2} + {{\mathbf{y}}^2} = {{\mathbf{m}}^2} + {{\mathbf{n}}^2}$

Ans: Taking RHS -

${\text{RHS}} = {{\mathbf{m}}^2} + {{\mathbf{n}}^2}$

$ = {\left( {{\text{x}}.{\text{cos A}} + {\text{y}}.{\text{sin A}}} \right)^2} + {\left( {{\text{x}}.{\text{sin A}} - {\text{y}}.{\text{cos A}}} \right)^2}$

$ = {{\text{x}}^2}.{\text{co}}{{\text{s}}^2}{\text{A}} + {{\text{y}}^2}{\text{si}}{{\text{n}}^2}{\text{A}} + 2{\text{xy}}.{\text{cos A}}.{\text{sin A}} +{{\text{x}}^2}.{\text{si}}{{\text{n}}^2}{\text{A}}+{{\text{y}}^2}{\text{co}}{{\text{s}}^2}{\text{A}} - 2{\text{xy}}.{\text{cos A}}.{\text{sin A}}$               $\left[ {{{\left( {{\mathbf{a}} \pm {\mathbf{b}}} \right)}^2} = {{\mathbf{a}}^2} + {{\mathbf{b}}^2} \pm 2{\mathbf{ab}}} \right]$

$ = {{\text{x}}^2}\left( {{\text{co}}{{\text{s}}^2}{\text{A}} + {\text{si}}{{\text{n}}^2}{\text{A}}} \right) + {{\mathbf{y}}^2}\left( {{\text{co}}{{\text{s}}^2}{\text{A}} + {\text{si}}{{\text{n}}^2}{\text{A}}} \right)$

$ = {{\mathbf{x}}^2} + {{\mathbf{y}}^2} = {\mathbf{LHS}}$                                       $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

3. If ${\mathbf{m}} = {\mathbf{a}}.{\mathbf{sec}}\;{\mathbf{A}} + {\mathbf{b}}.{\mathbf{tan}}\;{\mathbf{A}}$ and ${\mathbf{n}} = {\mathbf{a}}.{\mathbf{tan}}\;{\mathbf{A}} + {\mathbf{b}}.{\mathbf{sec}}\;{\mathbf{A}}$, then prove that: ${{\mathbf{m}}^2} - {{\mathbf{n}}^2} = {{\mathbf{a}}^2} - {{\mathbf{b}}^2}$

Ans: Taking LHS -

${\text{LHS}} = {{\mathbf{m}}^2} - {{\mathbf{n}}^2}$

$ = {\left( {{\mathbf{a}}.{\mathbf{sec}}\;{\mathbf{A}} + {\mathbf{b}}.{\mathbf{tan}}\;{\mathbf{A}}} \right)^2} - {\left( {{\mathbf{a}}.{\mathbf{tan}}\;{\mathbf{A}} + {\mathbf{b}}.{\mathbf{sec}}\;{\mathbf{A}}} \right)^2}$

$ = {{\text{a}}^2}.{\text{se}}{{\text{c}}^2}{\text{A}} + {{\text{b}}^2}{\text{ta}}{{\text{n}}^2}{\text{A}} + 2{\text{ab}}.{\text{sec A}}.{\text{tan A}} - {{\text{a}}^2}.{\text{ta}}{{\text{n}}^2}{\text{A}} - {{\text{b}}^2}{\text{se}}{{\text{c}}^2} - 2{\text{ab}}.{\text{sec A}}.{\text{tan A}}$

                                                                                  $\left[ {{{\left( {{\mathbf{a}} + {\mathbf{b}}} \right)}^2} = {{\mathbf{a}}^2} + {{\mathbf{b}}^2} + 2{\mathbf{ab}}} \right]$

$ = {{\text{a}}^2}\left( {{\text{se}}{{\text{c}}^2}{\text{A}} - {\text{ta}}{{\text{n}}^2}{\text{A}}} \right) - {{\mathbf{b}}^2}\left( {{\text{se}}{{\text{c}}^2}{\text{A}} - {\text{ta}}{{\text{n}}^2}{\text{A}}} \right)$

$ = {{\mathbf{a}}^2} - {{\mathbf{b}}^2} = {\mathbf{RHS}}$                                       $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;1 + {\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} = {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}} \right]$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

4. If ${\mathbf{x}} = {\mathbf{r}}\;{\mathbf{sin}}\;{\mathbf{A}}\;{\mathbf{cos}}\;{\mathbf{B}},\;{\mathbf{y}} = {\mathbf{r}}\;{\mathbf{sin}}\;{\mathbf{A}}\;{\mathbf{sin}}\;{\mathbf{B}}$ and ${\mathbf{z}} = {\mathbf{r}}\;{\mathbf{cos}}\;{\mathbf{A}}$, then prove that: ${{\mathbf{x}}^2} + {{\mathbf{y}}^2} + {{\mathbf{z}}^2} = {{\mathbf{r}}^2}$

Ans: Taking LHS -

${\text{LHS}} = {{\text{x}}^2} + {{\text{y}}^2} + {{\text{z}}^2}$

$ = {\left( {{\mathbf{r}}\;{\mathbf{sin}}\;{\mathbf{A}}\;{\mathbf{cos}}\;{\mathbf{B}}} \right)^2} + {\left( {{\mathbf{r}}\;{\mathbf{sin}}\;{\mathbf{A}}\;{\mathbf{sin}}\;{\mathbf{B}}} \right)^2} + {\left( {{\mathbf{r}}\;{\mathbf{cos}}\;{\mathbf{A}}} \right)^2}$

$ = {\left( {{\mathbf{r}}\;{\mathbf{sin}}\;{\mathbf{A}}} \right)^2}\left( {{\text{co}}{{\text{s}}^2}{\text{B}} + {\text{si}}{{\text{n}}^2}{\text{B}}} \right) + {\left( {{\mathbf{r}}\;{\mathbf{cos}}\;{\mathbf{A}}} \right)^2}$

                                                                     $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = {\left( {{\mathbf{r}}\;{\mathbf{sin}}\;{\mathbf{A}}} \right)^2} + {\left( {{\mathbf{r}}\;{\mathbf{cos}}\;{\mathbf{A}}} \right)^2}$

$ = {{\text{r}}^2}\left( {{\text{si}}{{\text{n}}^2}{\text{A}} + {\text{co}}{{\text{s}}^2}{\text{A}}} \right)$                                  $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = {{\mathbf{r}}^2} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

5. If ${\mathbf{sin}}\;{\mathbf{A}} + {\mathbf{cos}}\;{\mathbf{A}} = {\mathbf{m}}$ and ${\mathbf{sec}}\;{\mathbf{A}} + {\mathbf{cosec}}\;{\mathbf{A}} = {\mathbf{n}}$, show that: ${\mathbf{n}}\left( {{{\mathbf{m}}^2} - 1} \right) = 2{\mathbf{m}}$

Ans: Taking LHS -

${\text{LHS}} = {\text{n}}\left( {{{\text{m}}^2} - 1} \right)$

$ = \left( {{\text{sec\;A}} + {\text{cosec A}}} \right)\left[ {{{\left( {{\text{sin A}} + {\text{cos A}}} \right)}^2} - 1} \right]$

$ = \left( {{\text{sec\;A}} + {\text{cosec A}}} \right)\left[ {{\text{si}}{{\text{n}}^2}{\text{A}} + {\text{co}}{{\text{s}}^2}{\text{A}} + 2{\text{cos A}}.{\text{sin A}} - 1} \right]$

$\left[ {{{\left( {{\mathbf{a}} + {\mathbf{b}}} \right)}^2} = {{\mathbf{a}}^2} + {{\mathbf{b}}^2} + 2{\mathbf{ab}}} \right]$

$ = \left( {{\text{sec\;A}} + {\text{cosec\;A}}} \right)\left[ {1 + 2{\text{cos A}}.{\text{sin A}} - 1} \right]$ $\left[{{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \left( {{\text{sec A}} + {\text{cosec A}}} \right)\left[ {2{\text{cos A}}.{\text{sin A}}} \right]$

$ = \left( {{\text{sec A}}} \right)\left[ {2{\text{cos A}}.{\text{sin A}}} \right] + {\text{cosec A}}\left[ {2{\text{cos A}}.{\text{sin A}}} \right]$

$ = \left[ {2{\text{sin A}}} \right] + \left[ {2{\text{cos A}}} \right]$            $\left[ {{\text{We know}}:{\text{cos A}}.{\text{sec A}} = 1,{\text{cosec A}}.{\text{sin A}} = 1} \right]$

$ = 2\left[ {{\text{sin A}} + {\text{cos A}}} \right]$

$ = 2{\mathbf{m}} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

6. If ${\mathbf{x}} = {\mathbf{r}}\;{\mathbf{cos}}\;{\mathbf{A}}\;{\mathbf{cos}}\;{\mathbf{B}},\;{\mathbf{y}} = {\mathbf{r}}\;{\mathbf{cos}}\;{\mathbf{A}}\;{\mathbf{sin}}\;{\mathbf{B}}$ and ${\mathbf{z}} = {\mathbf{r}}\;{\mathbf{sin}}\;{\mathbf{A}}$, show that: ${{\mathbf{x}}^2} + {{\mathbf{y}}^2} + {{\mathbf{z}}^2} = {{\mathbf{r}}^2}$

Ans: Taking LHS -

${\text{LHS}} = {{\text{x}}^2} + {{\text{y}}^2} + {{\text{z}}^2}$

$ = {\left( {{\mathbf{r}}\;{\mathbf{cos}}\;{\mathbf{A}}\;{\mathbf{cos}}\;{\mathbf{B}}} \right)^2} + {\left( {{\mathbf{r}}\;{\mathbf{cos}}\;{\mathbf{A}}\;{\mathbf{sin}}\;{\mathbf{B}}} \right)^2} + {\left( {{\mathbf{r}}\;{\mathbf{sin}}\;{\mathbf{A}}} \right)^2}$

$ = {\left( {{\mathbf{r}}\;{\mathbf{cos}}\;{\mathbf{A}}} \right)^2}\left( {{\text{co}}{{\text{s}}^2}{\text{B}} + {\text{si}}{{\text{n}}^2}{\text{B}}} \right) + {\left( {{\mathbf{r}}\;{\mathbf{sin}}\;{\mathbf{A}}} \right)^2}$ $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = {\left( {{\mathbf{r}}\;{\mathbf{cos}}\;{\mathbf{A}}} \right)^2} + {\left( {{\mathbf{r}}\;{\mathbf{sin}}\;{\mathbf{A}}} \right)^2}$

$ = {{\text{r}}^2}\left( {{\text{co}}{{\text{s}}^2}{\text{A}} + {\text{si}}{{\text{n}}^2}{\text{A}}} \right)$                                  $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = {{\mathbf{r}}^2} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

7. If $\dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{B}}}} = {\mathbf{m}}$ and $\dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{B}}}} = {\mathbf{n}}$, show that: $\left( {\;{{\mathbf{m}}^2} + {{\mathbf{n}}^2}\;} \right){\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}} = {{\mathbf{n}}^2}.$

Ans: Taking LHS -

${\text{LHS}} = \left( {{\text{}}{{\text{m}}^2} + {{\text{n}}^2}{\text{}}} \right){\text{co}}{{\text{s}}^2}{\text{B}}$

$ = \left[ {{{\left( {\dfrac{{{\text{cos A}}}}{{{\text{cos B}}}}} \right)}^2} + {{\left( {\dfrac{{{\text{cos A}}}}{{{\text{sin B}}}}} \right)}^2}} \right]{\text{co}}{{\text{s}}^2}{\text{B}}$

$ = \left[ {{{\left( {\dfrac{1}{{{\text{cos B}}}}} \right)}^2} + {{\left( {\dfrac{1}{{{\text{sin B}}}}} \right)}^2}} \right]{\text{co}}{{\text{s}}^2}{\text{A}}.{\text{co}}{{\text{s}}^2}{\text{B}}$

$ = \left[ {{{\left( {\dfrac{{{\text{si}}{{\text{n}}^2}{\text{B}} + {\text{co}}{{\text{s}}^2}{\text{B}}}}{{{\text{co}}{{\text{s}}^2}{\text{B}}.{\text{si}}{{\text{n}}^2}{\text{B}}}}} \right)}^{}}} \right]{\text{co}}{{\text{s}}^2}{\text{A}}.{\text{co}}{{\text{s}}^2}{\text{B}}$

$ = \left[ {{{\left( {\dfrac{1}{{{\text{co}}{{\text{s}}^2}{\text{B}}.{\text{si}}{{\text{n}}^2}{\text{B}}}}} \right)}^{}}} \right]{\text{co}}{{\text{s}}^2}{\text{A}}.{\text{co}}{{\text{s}}^2}{\text{B}}$              $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \left[ {{{\left( {\dfrac{1}{{{\text{si}}{{\text{n}}^2}{\text{B}}}}} \right)}^{}}} \right]{\text{co}}{{\text{s}}^2}{\text{A}}$

$ = {\left( {\dfrac{{{\text{cos A}}}}{{{\text{sin\;B}}}}} \right)^2} = {{\mathbf{n}}^2} = {\mathbf{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

Exercise - 21(C)

1. Show that: 

i) ${\mathbf{tan}}\;{10^ \circ }{\mathbf{tan}}\;{15^ \circ }{\mathbf{tan}}\;{75^ \circ }{\mathbf{tan}}\;{80^ \circ } = 1$

Ans: Taking LHS -

${\text{LHS}} = {\text{tan}}{10^ \circ }{\text{tan}}{15^ \circ }{\text{tan}}{75^ \circ }{\text{tan}}{80^ \circ }$

$ = {\text{tan}}{\left( {90 - 80} \right)^ \circ }{\text{tan}}{\left( {90 - 75} \right)^ \circ }{\text{tan}}{75^ \circ }{\text{tan}}{80^ \circ }$

$ = {\text{cot}}{80^ \circ }{\text{cot}}{75^ \circ }{\text{tan}}{75^ \circ }{\text{tan}}{80^ \circ }$     $\left[ {{\text{We know that}}:{\text{tan}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{cot A}}} \right]$

$ = 1 = {\mathbf{RHS}}$                                             $\left[ {{\text{We know that}}:{\text{tan A}}.{\text{cot A}} = 1} \right]$             

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

ii) ${\mathbf{sin}}\;{42^ \circ }{\mathbf{sec}}\;{48^ \circ } + {\mathbf{cos}}\;{42^ \circ }{\mathbf{cosec}}\;{48^ \circ } = 2$

Ans: Taking LHS -

${\text{LHS}} = {\text{sin}}{42^ \circ }{\text{sec}}{48^ \circ } + {\text{cos}}{42^ \circ }{\text{cosec}}{48^ \circ }$

$ = {\text{sin}}{42^ \circ }{\text{sec}}{\left( {90 - 42} \right)^ \circ } + {\text{cos}}{42^ \circ }{\text{cosec}}{(90 - 42)^ \circ }$

$ = {\text{sin}}{42^ \circ }{\text{cosec}}{42^ \circ } + {\text{cos}}{42^ \circ }{\text{sec}}{42^ \circ }$

$\left[ {{\text{We know that}}:{\text{sec}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{cosec A}},{\text{cosec}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{sec A}}} \right]$

$ = 1 + 1$                       $\left[ {{\text{We know that}}:{\text{sin A}}.{\text{cosec A}} = 1,{\text{cos A}}.{\text{sec A}} = 1} \right]$  

$ = 2 = {\mathbf{RHS}}$                                                        

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

iii) $\dfrac{{{\mathbf{sin}}\;{{26}^ \circ }}}{{{\mathbf{sec}}\;{{64}^ \circ }}} + \dfrac{{{\mathbf{cos}}\;{{26}^ \circ }}}{{{\mathbf{cosec}}\;{{64}^ \circ }}} = 1$

Ans: Taking LHS -

${\text{LHS}} = \dfrac{{{\text{sin}}{{26}^ \circ }}}{{{\text{sec}}{{64}^ \circ }}} + \dfrac{{{\text{cos}}{{26}^ \circ }}}{{{\text{cosec}}{{64}^ \circ }}}$

$ = \dfrac{{{\text{sin}}{{26}^ \circ }}}{{{\text{sec}}{{(90{\text{}} - {\text{}}26)}^ \circ }}} + \dfrac{{{\text{cos}}{{26}^ \circ }}}{{{\text{cosec}}{{(90{\text{}} - {\text{}}26)}^ \circ }}}$

$ = \dfrac{{{\text{sin}}{{26}^ \circ }}}{{{\text{cosec}}{{(26)}^ \circ }}} + \dfrac{{{\text{cos}}{{26}^ \circ }}}{{{\text{sec}}{{(26)}^ \circ }}}$

$\left[ {{\text{We know that}}:{\text{sec}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{cosec A}},{\text{cosec}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{sec A}}} \right]$

$ = {\text{si}}{{\text{n}}^2}{26^ \circ } + {\text{co}}{{\text{s}}^2}{26^ \circ }$           $\left[ {{\text{We know that}}:{\text{cosec A}} = \dfrac{1}{{{\text{sin A}}}},{\text{sec A}} = \dfrac{1}{{{\text{cos A}}}}} \right]$  

$ = 1 = {\mathbf{RHS}}$                                                  $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$                      

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

2. Express each of the following in terms of angles between ${0^ \circ }$and ${45^ \circ }$:

i) ${\mathbf{sin}}\;{59^ \circ } + {\mathbf{tan}}\;{63^ \circ }$

Ans: ${\mathbf{sin}}\;{59^ \circ } + {\mathbf{tan}}\;{63^ \circ } = {\mathbf{sin}}{\left( {90 - 31} \right)^ \circ } + {\mathbf{tan}}{\left( {90 - 27} \right)^ \circ }$

$\left[ {{\text{We know that}}:{\text{sin}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{cos A}},{\text{tan}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{cot A}}} \right]$

$ \Rightarrow {\mathbf{sin}}\;{59^ \circ } + {\mathbf{tan}}\;{63^ \circ } = {\mathbf{cos}}\;{31^ \circ } + {\mathbf{cot}}\;{27^ \circ }$

ii) ${\mathbf{cosec}}\;{68^ \circ } + {\mathbf{cot}}\;{72^ \circ }$

Ans: ${\mathbf{cosec}}\;{68^ \circ } + {\mathbf{cot}}\;{72^ \circ } = {\mathbf{cosec}}{\left( {90 - 22} \right)^ \circ } + {\mathbf{cot}}{\left( {90 - 18} \right)^ \circ }$

$\left[ {{\text{We know that}}:{\text{cosec}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{sec A}},{\text{cot}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{tan A}}} \right]$

$ \Rightarrow {\mathbf{cosec}}\;{68^ \circ } + {\mathbf{cot}}\;{72^ \circ } = {\mathbf{sec}}\;{22^ \circ } + {\mathbf{tan}}\;{18^ \circ }$

iii) ${\mathbf{cos}}\;{74^ \circ } + {\mathbf{sec}}\;{67^ \circ }$

Ans: ${\mathbf{cos}}\;{74^ \circ } + {\mathbf{sec}}\;{67^ \circ } = {\mathbf{cos}}{\left( {90 - 16} \right)^ \circ } + {\mathbf{sec}}{\left( {90 - 23} \right)^ \circ }$

$\left[ {{\text{We know that}}:{\text{cos}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{sin A}},{\text{sec}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{cosec A}}} \right]$

$ \Rightarrow {\mathbf{cos}}\;{74^ \circ } + {\mathbf{sec}}\;{67^ \circ } = {\mathbf{sin}}\;{16^ \circ } + {\mathbf{cosec}}\;{23^ \circ }$

3. Show that:

i) $\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;\left( {90\; - \;{\mathbf{A}}} \right)}} + \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;\left( {90\; - \;{\mathbf{A}}} \right)}} = {\mathbf{sec}}\;{\mathbf{A}}\;.\;{\mathbf{cosec}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\text{LHS}} = \dfrac{{{\text{sin A}}}}{{{\text{sin}}\left( {90{\text{}} - {\text{A}}} \right)}} + \dfrac{{{\text{cos A}}}}{{{\text{cos}}\left( {90{\text{}} - {\text{A}}} \right)}}$

$ = \dfrac{{{\text{sin A}}}}{{{\text{cos A}}}} + \dfrac{{{\text{cos A}}}}{{{\text{sin A}}}}$                       $\left[ {{\text{cos}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{sin A}},{\text{sin}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{cos A}}} \right]$

$ = \dfrac{{{\text{si}}{{\text{n}}^2}{\text{A}} + {\text{co}}{{\text{s}}^2}{\text{A}}}}{{{\text{cos A}}.{\text{sin A}}}}$

$ = \dfrac{1}{{{\text{cos A}}.{\text{sin A}}}}$                                                  $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = {\mathbf{sec}}\;{\mathbf{A}}\;.\;{\mathbf{cosec}}\;{\mathbf{A}} = {\mathbf{RHS}}$      $\left[ {{\text{We know that}}:{\text{cosec A}} = \dfrac{1}{{{\text{sin A}}}},{\text{sec A}} = \dfrac{1}{{{\text{cos A}}}}} \right]$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

ii) ${\mathbf{sin}}\;{\mathbf{A}}\;{\mathbf{cos}}\;{\mathbf{A}} - \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\;{\mathbf{cos}}\left( {90\; - \;{\mathbf{A}}} \right)\;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sec}}\;\left( {90\; - \;{\mathbf{A}}} \right)}} - \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\;{\mathbf{sin}}\left( {90\; - \;{\mathbf{A}}} \right)\;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{cosec}}\;\left( {90\; - \;{\mathbf{A}}} \right)}} = 0$

Ans: Taking LHS -

${\text{LHS}} = {\text{sin A. cos A}} - \dfrac{{{\text{sin A.cos}}\left( {90{\text{}} - {\text{A}}} \right){\text{cos A}}}}{{{\text{sec}}\left( {90{\text{}} - {\text{A}}} \right)}} - \dfrac{{{\text{sin A sin}}\left( {90{\text{}} - {\text{A}}} \right){\text{cos A}}}}{{{\text{cosec}}\left( {90{\text{}} - {\text{A}}} \right)}}$

$ = {\text{sin A.cos A}} - \dfrac{{{\text{sin A sin A cos A}}}}{{{\text{cosec A}}}} - \dfrac{{{\text{sin A.cos A cos A}}}}{{{\text{sec A}}}}$ $\left[ {{\text{We know that}}:{\text{cos}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{sin A}},{\text{sin}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{cos A}}} \right]$

$\left[ {{\text{We know that}}:{\text{cosec}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{sec A}},{\text{sec}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{cosec A}}} \right]$

$ = {\text{sin A.cos A}} - {\text{si}}{{\text{n}}^3}{\text{A cos A}} - {\text{co}}{{\text{s}}^3}{\text{A sin A}}$

$\left[ {{\text{We know that}}:{\text{cosecA}} = \dfrac{1}{{{\text{sinA}}}},{\text{sec A}} = \dfrac{1}{{{\text{cosA}}}}} \right]$

$ = {\text{sinA.cosA}} - {\text{sinA.cosA}}\left( {{\text{si}}{{\text{n}}^2}{\text{A}} + {\text{co}}{{\text{s}}^2}{\text{A}}} \right)$

$\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = {\text{sinA.cosA}} - {\text{sinA.cosA}}\left( 1 \right)$

$ = {\text{sinA.cosA}} - {\text{sinA.cosA}}$

$ = 0 = {\text{RHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

4. For triangle ABC, show that:

i) ${\mathbf{sin}}\;\left( {\dfrac{{{\mathbf{A}}\; + \;{\mathbf{B}}}}{2}} \right) = {\mathbf{cos}}\;\dfrac{{\mathbf{C}}}{2}$

Ans: ${\mathbf{sin}}\;\left( {\dfrac{{{\mathbf{A}}\; + \;{\mathbf{B}}}}{2}} \right)$

We know that for a ${{\Delta ABC}},$

$\angle {\mathbf{A}} + \angle {\mathbf{B}} + \angle {\mathbf{C}} = {180^ \circ }$

$ \Rightarrow \dfrac{{\angle {\mathbf{A}}\; + \;\angle {\mathbf{B}}\; + \angle {\mathbf{C}}}}{2} = \dfrac{{{{180}^ \circ }}}{2}$

$ \Rightarrow \dfrac{{\angle {\mathbf{A}}\; + \;\angle {\mathbf{B}}\; + \angle {\mathbf{C}}}}{2} = {90^ \circ }^{}$

$ \Rightarrow \dfrac{{\angle {\mathbf{A}}\; + \;\angle {\mathbf{B}}}}{2} = {90^ \circ } - {\dfrac{{\angle {\mathbf{C}}}}{2}^{}}$

${\mathbf{sin}}\;\left( {\dfrac{{{\mathbf{A}}\; + \;{\mathbf{B}}}}{2}} \right) = {\mathbf{sin}}\;\left( {{{90}^ \circ } - \dfrac{{\mathbf{C}}}{2}} \right)$

$ \Rightarrow {\mathbf{sin}}\;\left( {\dfrac{{{\mathbf{A}}\; + \;{\mathbf{B}}}}{2}} \right) = {\mathbf{cos}}\;\left( {\dfrac{{\mathbf{C}}}{2}} \right)$             $\left[ {{\text{We know that}}:{\text{sin}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{cos A}}} \right]$

Hence proved.

ii) ${\mathbf{tan}}\;\left( {\dfrac{{{\mathbf{B}}\; + \;{\mathbf{C}}}}{2}} \right) = {\mathbf{cos}}\;\dfrac{{\mathbf{A}}}{2}$

Ans: ${\mathbf{tan}}\;\left( {\dfrac{{{\mathbf{B}}\; + \;{\mathbf{C}}}}{2}} \right)$

We know that for a ${\text{\Delta ABC}},$

$\angle {\mathbf{A}} + \angle {\mathbf{B}} + \angle {\mathbf{C}} = {180^ \circ }$

$ \Rightarrow \dfrac{{\angle {\mathbf{A}}\; + \;\angle {\mathbf{B}}\; + \angle {\mathbf{C}}}}{2} = \dfrac{{{{180}^ \circ }}}{2}$

$ \Rightarrow \dfrac{{\angle {\mathbf{A}}\; + \;\angle {\mathbf{B}}\; + \angle {\mathbf{C}}}}{2}$ = 90o

$ \Rightarrow \dfrac{{\angle {\mathbf{B}}\; + \;\angle {\mathbf{C}}}}{2} = $ 90o - $\dfrac{{\angle {\mathbf{A}}}}{2}$

${\mathbf{tan}}\;\left( {\dfrac{{{\mathbf{B}}\; + \;{\mathbf{C}}}}{2}} \right) = {\mathbf{tan}}\;\left( {{{90}^ \circ } - \dfrac{{\mathbf{A}}}{2}} \right)$

$ \Rightarrow {\mathbf{tan}}\;\left( {\dfrac{{{\mathbf{B}}\; + \;{\mathbf{C}}}}{2}} \right) = {\mathbf{cot}}\;\left( {\dfrac{{\mathbf{A}}}{2}} \right)$             $\left[ {{\text{We know that}}:{\text{tan}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{cot A}}} \right]$

Hence proved.

5. Evaluate:

i) $3\dfrac{{{\mathbf{sin}}\;{{72}^ \circ }}}{{{\mathbf{cos}}\;{{18}^ \circ }}} - \dfrac{{{\mathbf{sec}}\;{{32}^ \circ }}}{{{\mathbf{cosec}}\;{{58}^ \circ }}}$

Ans: $3\dfrac{{{\mathbf{sin}}\;{{72}^ \circ }}}{{{\mathbf{cos}}\;{{18}^ \circ }}} - \dfrac{{{\mathbf{sec}}\;{{32}^ \circ }}}{{{\mathbf{cosec}}\;{{58}^ \circ }}}$

$ = 3\dfrac{{{\mathbf{sin}}\;{{72}^ \circ }}}{{{\mathbf{cos}}\;{{\left( {90\; - \;72} \right)}^ \circ }}} - \dfrac{{{\mathbf{sec}}\;{{32}^ \circ }}}{{{\mathbf{cosec}}\;{{\left( {90\; - \;32} \right)}^ \circ }}}$

$\left[ {{\text{We know that}}:{\text{cos}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{sinA}},{\text{cosec}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{secA}}} \right]$

$ = 3\dfrac{{{\mathbf{sin}}\;{{72}^ \circ }}}{{{\mathbf{sin}}\;{{72}^ \circ }}} - \dfrac{{{\mathbf{sec}}\;{{32}^ \circ }}}{{{\mathbf{sec}}\;{{32}^ \circ }}}$ 

$ = 3 \times 1 - 1$

$ = 3 - 1$

$ = 2$

ii) $3{\mathbf{cos}}\;{80^ \circ }{\mathbf{cosec}}\;{10^ \circ } + 2{\mathbf{sin}}\;{59^ \circ }{\mathbf{sec}}\;{31^ \circ }$

Ans: $3{\mathbf{cos}}\;{80^ \circ }{\mathbf{cosec}}\;{10^ \circ } + 2{\mathbf{sin}}\;{59^ \circ }{\mathbf{sec}}\;{31^ \circ }$

$ = 3\dfrac{{{\mathbf{cos}}\;{{80}^ \circ }}}{{{\mathbf{sin}}\;{{10}^ \circ }}} + 2\dfrac{{{\mathbf{sin}}\;{{59}^ \circ }}}{{{\mathbf{cos}}\;{{31}^ \circ }}}$           $\left[ {{\text{We know that}}:{\text{cosecA}} = \dfrac{1}{{{\text{sin A}}}},{\text{secA}} = \dfrac{1}{{{\text{cosA}}}}} \right]$

$ = 3\dfrac{{{\mathbf{cos}}\;{{\left( {90\; - \;10} \right)}^ \circ }}}{{{\mathbf{sin}}\;{{10}^ \circ }}} + 2\dfrac{{{\mathbf{sin}}\;{{\left( {90\; - \;31} \right)}^ \circ }}}{{{\mathbf{cos}}\;{{31}^ \circ }}}$ $\left[ {{\text{We know that}}:{\text{cos}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{sinA}},{\text{sin}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{cosA}}} \right]$

$ = 3\dfrac{{{\mathbf{sin}}\;{{10}^ \circ }}}{{{\mathbf{sin}}\;{{10}^ \circ }}} + 2\dfrac{{{\mathbf{cos}}\;{{31}^ \circ }}}{{{\mathbf{cos}}\;{{31}^ \circ }}}$ 

$ = 3 \times 1 + \left( {2 \times 1} \right)$

$ = 3 + 2$

$ = 5$

iii) $\dfrac{{{\mathbf{sin}}\;{{80}^ \circ }}}{{{\mathbf{cos}}\;{{10}^ \circ }}} + {\mathbf{sin}}\;{59^ \circ }{\mathbf{sec}}\;{31^ \circ }$

Ans: $\dfrac{{{\mathbf{sin}}\;{{80}^ \circ }}}{{{\mathbf{cos}}\;{{10}^ \circ }}} + {\mathbf{sin}}\;{59^ \circ }{\mathbf{sec}}\;{31^ \circ }$

$ = \dfrac{{{\mathbf{sin}}\;{{80}^ \circ }}}{{{\mathbf{cos}}\;{{\left( {90\; - \;80} \right)}^ \circ }}} + \dfrac{{{\mathbf{sin}}\;{{59}^ \circ }}}{{{\mathbf{cos}}\;{{31}^ \circ }}}$                                       $\left[ {{\text{We know that}}:{\text{sec A}} = \dfrac{1}{{{\text{cos A}}}}} \right]$

$ = \dfrac{{{\mathbf{sin}}\;{{80}^ \circ }}}{{{\mathbf{sin}}\;{{80}^ \circ }}} + \dfrac{{{\mathbf{sin}}\;{{59}^ \circ }}}{{{\mathbf{cos}}\;{{\left( {90\; - \;59} \right)}^ \circ }}}$                        $\left[ {{\text{We know that}}:{\text{cos}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{sin A}}} \right]$

$ = 1 + \dfrac{{{\mathbf{sin}}\;{{59}^ \circ }}}{{{\mathbf{sin}}\;{{59}^ \circ }}}$                                         $\left[ {{\text{We know that}}:{\text{cos}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{sin A}}} \right]$

$ = 1 + 1$

$ = 2$

iv) ${\mathbf{tan}}\left( {{{55}^ \circ } - {\mathbf{A}}} \right) - {\mathbf{cot}}\left( {{{35}^ \circ } + {\mathbf{A}}} \right)$

Ans: ${\mathbf{tan}}\left( {{{55}^ \circ } - {\mathbf{A}}} \right) - {\mathbf{cot}}\left( {{{35}^ \circ } + {\mathbf{A}}} \right)$

$ = {\mathbf{tan}}\left[ {{{90}^ \circ } - \left( {{{35}^ \circ } + {\mathbf{A}}} \right)} \right] - {\mathbf{cot}}\left( {{{35}^ \circ } + {\mathbf{A}}} \right)$

$ = {\mathbf{cot}}\left( {{{35}^ \circ } + {\mathbf{A}}} \right) - {\mathbf{cot}}\left( {{{35}^ \circ } + {\mathbf{A}}} \right)$     $\left[ {{\text{We know that}}:{\text{tan}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{cot A}}} \right]$

$ = 0$

v) ${\mathbf{cosec}}\left( {{{65}^ \circ } + {\mathbf{A}}} \right) - {\mathbf{sec}}\left( {{{25}^ \circ } - {\mathbf{A}}} \right)$

Ans: ${\mathbf{cosec}}\left( {{{65}^ \circ } + {\mathbf{A}}} \right) - {\mathbf{sec}}\left( {{{25}^ \circ } - {\mathbf{A}}} \right)$

$ = {\mathbf{cosec}}\left[ {{{90}^ \circ } - \left( {{{25}^ \circ } - {\mathbf{A}}} \right)} \right] - {\mathbf{sec}}\left( {{{25}^ \circ } - {\mathbf{A}}} \right)$

$ = {\mathbf{sec}}\left( {{{25}^ \circ } - {\mathbf{A}}} \right) - {\mathbf{sec}}\left( {{{25}^ \circ } - {\mathbf{A}}} \right)$  $\left[ {{\text{We know that}}:{\text{cosec}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{sec A}}} \right]$

$ = 0$

vi) $2\dfrac{{{\mathbf{tan}}\;{{57}^ \circ }}}{{{\mathbf{cot}}\;{{33}^ \circ }}} - \dfrac{{{\mathbf{cot}}\;{{70}^ \circ }}}{{{\mathbf{tan}}\;{{20}^ \circ }}} - \sqrt 2 {\mathbf{cos}}\;{45^ \circ }$

Ans: $2\dfrac{{{\mathbf{tan}}\;{{57}^ \circ }}}{{{\mathbf{cot}}\;{{33}^ \circ }}} - \dfrac{{{\mathbf{cot}}\;{{70}^ \circ }}}{{{\mathbf{tan}}\;{{20}^ \circ }}} - \sqrt 2 {\mathbf{cos}}\;{45^ \circ }$

$ = 2\dfrac{{{\mathbf{tan}}\;{{57}^ \circ }}}{{{\mathbf{cot}}\;{{\left( {90\; - \;57} \right)}^ \circ }}} - \dfrac{{{\mathbf{cot}}\;{{\left( {90\; - \;20} \right)}^ \circ }}}{{{\mathbf{tan}}\;{{20}^ \circ }}} - \sqrt 2  \times \dfrac{1}{{\sqrt 2 }}$       $\left[ {{\text{We know that}}:{\mathbf{cos}}\;{{45}^ \circ } = \dfrac{1}{{\sqrt 2 }}} \right]$

$ = 2\dfrac{{{\mathbf{tan}}\;{{57}^ \circ }}}{{{\mathbf{tan}}\;{{57}^ \circ }}} - \dfrac{{{\mathbf{tan}}\;{{20}^ \circ }}}{{{\mathbf{tan}}\;{{20}^ \circ }}} - 1$                     $\left[ {{\text{We know that}}:{\text{cot}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{tan A}}} \right]$

$ = 2 \times 1 - 1 - 1$

$ = 0$

vii) $\dfrac{{{\mathbf{co}}{{\mathbf{t}}^2}{{41}^ \circ }}}{{{\mathbf{ta}}{{\mathbf{n}}^2}{{49}^ \circ }}} - 2\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{{75}^ \circ }}}{{{\mathbf{co}}{{\mathbf{s}}^2}{{15}^ \circ }}}$

Ans: $\dfrac{{{\mathbf{co}}{{\mathbf{t}}^2}{{41}^ \circ }}}{{{\mathbf{ta}}{{\mathbf{n}}^2}{{49}^ \circ }}} - 2\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{{75}^ \circ }}}{{{\mathbf{co}}{{\mathbf{s}}^2}{{15}^ \circ }}}$

$ = \dfrac{{{\mathbf{co}}{{\mathbf{t}}^2}{{41}^ \circ }}}{{{\mathbf{ta}}{{\mathbf{n}}^2}{{\left( {90\; - \;41} \right)}^ \circ }}} - 2\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{{75}^ \circ }}}{{{\mathbf{co}}{{\mathbf{s}}^2}{{\left( {90\; - \;75} \right)}^ \circ }}}$     

$ = \dfrac{{{\mathbf{co}}{{\mathbf{t}}^2}{{41}^ \circ }}}{{{\mathbf{co}}{{\mathbf{t}}^2}{{41}^ \circ }}} - 2\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{{75}^ \circ }}}{{{\mathbf{si}}{{\mathbf{n}}^2}{{75}^ \circ }}}$     $\left[ {{\text{We know that}}:{\text{tan}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{cot A}},{\text{cos}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{sin A}}} \right]$

$ = 1 - \left( {2 \times 1} \right)$

$ =  - 1$

viii) $\dfrac{{{\mathbf{cos}}\;{{70}^ \circ }}}{{{\mathbf{sin}}\;{{20}^ \circ }}} + \dfrac{{{\mathbf{cos}}\;{{59}^ \circ }}}{{{\mathbf{sin}}\;{{31}^ \circ }}} - 8{\mathbf{si}}{{\mathbf{n}}^2}{30^ \circ }$

Ans: $\dfrac{{{\mathbf{cos}}\;{{70}^ \circ }}}{{{\mathbf{sin}}\;{{20}^ \circ }}} + \dfrac{{{\mathbf{cos}}\;{{59}^ \circ }}}{{{\mathbf{sin}}\;{{31}^ \circ }}} - 8{\mathbf{si}}{{\mathbf{n}}^2}{30^ \circ }$

$ = \dfrac{{{\mathbf{cos}}\;{{70}^ \circ }}}{{{\mathbf{sin}}\;{{\left( {90\; - \;70} \right)}^ \circ }}} + \dfrac{{{\mathbf{cos}}\;{{59}^ \circ }}}{{{\mathbf{sin}}\;{{\left( {90\; - \;59} \right)}^ \circ }}} - 8{\left( {\dfrac{1}{2}} \right)^2}$               $\left[ {{\text{We know that}}:{\mathbf{sin}}\;{{30}^ \circ } = \dfrac{1}{2}} \right]$

$ = \dfrac{{{\mathbf{cos}}\;{{70}^ \circ }}}{{{\mathbf{cos}}\;{{70}^ \circ }}} + \dfrac{{{\mathbf{cos}}\;{{59}^ \circ }}}{{{\mathbf{cos}}\;{{59}^ \circ }}} - 8 \times \dfrac{1}{4}$              $\left[ {{\text{We know that}}:{\text{sin}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{cos A}}} \right]$

$ = 1 + 1 - 2$

$ = 0$

ix) $14{\mathbf{sin}}\;{30^ \circ } + 6{\mathbf{cos}}\;{60^ \circ } - 5{\mathbf{tan}}\;{45^ \circ }$

Ans: $14{\mathbf{sin}}\;{30^ \circ } + 6{\mathbf{cos}}\;{60^ \circ } - 5{\mathbf{tan}}\;{45^ \circ }$

$ = \left( {14 \times \dfrac{1}{2}} \right) + \left( {6 \times \dfrac{1}{2}} \right) - \left( {5 \times 1} \right)$               $\left[ {{\text{We know that}}:{\mathbf{sin}}\;{{30}^ \circ } = {\mathbf{cos}}\;{{60}^ \circ } = \dfrac{1}{2},\;{\mathbf{tan}}\;{{45}^ \circ } = 1} \right]$

$ = 7 + 3 - 5$

$ = 5$

6. A triangle ABC is right angled at B; find the value of $\dfrac{{{\mathbf{sec}}\;{\mathbf{A}}.\;{\mathbf{cosec}}\;{\mathbf{C}}\; - \;{\mathbf{tan}}\;{\mathbf{A}}.\;{\mathbf{cot}}\;{\mathbf{C}}}}{{{\mathbf{sin}}\;{\mathbf{B}}}}.$

Ans: $\dfrac{{{\mathbf{sec}}\;{\mathbf{A}}.\;{\mathbf{cosec}}\;{\mathbf{C}}\; - \;{\mathbf{tan}}\;{\mathbf{A}}.\;{\mathbf{cot}}\;{\mathbf{C}}}}{{{\mathbf{sin}}\;{\mathbf{B}}}}$

Since, triangle ABC is right angle triangle and right angled at B.

$ \Rightarrow \angle {\text{B}} = {90^ \circ }$

We know that $\angle {\text{A}} + \angle {\text{B}} + \angle {\text{C}} = {180^ \circ }$

$ \Rightarrow \angle {\text{A}} + \angle {\text{C}} = {90^ \circ }$

$\dfrac{{{\mathbf{sec}}\;{\mathbf{A}}.\;{\mathbf{cosec}}\;{\mathbf{C}}\; - \;{\mathbf{tan}}\;{\mathbf{A}}.\;{\mathbf{cot}}\;{\mathbf{C}}}}{{{\mathbf{sin}}\;{\mathbf{B}}}} = \dfrac{{{\mathbf{sec}}\;\left( {90\; - \;{\mathbf{C}}} \right)\;.\;{\mathbf{cosec}}\;{\mathbf{C}}\; - \;{\mathbf{tan}}\;\left( {90\; - \;{\mathbf{C}}} \right)\;.\;{\mathbf{cot}}\;{\mathbf{C}}}}{{{\mathbf{sin}}\;{{90}^ \circ }}}$

$ \Rightarrow \dfrac{{{\mathbf{sec}}\;{\mathbf{A}}.\;{\mathbf{cosec}}\;{\mathbf{C}}\; - \;{\mathbf{tan}}\;{\mathbf{A}}.\;{\mathbf{cot}}\;{\mathbf{C}}}}{{{\mathbf{sin}}\;{\mathbf{B}}}} = \dfrac{{{\mathbf{cosec}}\;{\mathbf{C}}\;.\;{\mathbf{cosec}}\;{\mathbf{C}}\; - \;{\mathbf{cot}}\;{\mathbf{C}}\;.\;{\mathbf{cot}}\;{\mathbf{C}}}}{1}$

$\left[ {{\text{We know that}}:{\text{sec}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{cosec A}},{\text{tan}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{cot A}}} \right]$

$ \Rightarrow \dfrac{{{\mathbf{sec}}\;{\mathbf{A}}.\;{\mathbf{cosec}}\;{\mathbf{C}}\; - \;{\mathbf{tan}}\;{\mathbf{A}}.\;{\mathbf{cot}}\;{\mathbf{C}}}}{{{\mathbf{sin}}\;{\mathbf{B}}}} = {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{C}}\; - \;{\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{C}}$

$ \Rightarrow \dfrac{{{\mathbf{sec}}\;{\mathbf{A}}.\;{\mathbf{cosec}}\;{\mathbf{C}}\; - \;{\mathbf{tan}}\;{\mathbf{A}}.\;{\mathbf{cot}}\;{\mathbf{C}}}}{{{\mathbf{sin}}\;{\mathbf{B}}}} = 1$               $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{C}}\; - \;{\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{C}} = 1} \right]$

7. Find ( in each case, given below) the value of ${\mathbf{x}}$ if:

i) ${\mathbf{sin}}\;{\mathbf{x}} = {\mathbf{sin}}\;{60^ \circ }.{\mathbf{cos}}\;{30^ \circ } - {\mathbf{cos}}\;{60^ \circ }.{\mathbf{sin}}\;{30^ \circ }$

Ans: ${\mathbf{sin}}\;{\mathbf{x}} = {\mathbf{sin}}\;{60^ \circ }.{\mathbf{cos}}\;{30^ \circ } - {\mathbf{cos}}\;{60^ \circ }.{\mathbf{sin}}\;{30^ \circ }$

$\left[ {{\text{We know that}}:{\mathbf{sin}}\;{{30}^ \circ } = {\mathbf{cos}}\;{{60}^ \circ } = \dfrac{1}{2},\;{\mathbf{sin}}\;{{60}^ \circ } = {\mathbf{cos}}\;{{30}^ \circ } = \dfrac{{\sqrt 3 }}{2}} \right]$

$ \Rightarrow {\mathbf{sin}}\;{\mathbf{x}} = \left( {\dfrac{{\sqrt 3 }}{2}} \right).\left( {\dfrac{{\sqrt 3 }}{2}} \right) - \left( {\dfrac{1}{2}} \right).\left( {\dfrac{1}{2}} \right)$

$ \Rightarrow {\mathbf{sin}}\;{\mathbf{x}} = \left( {\dfrac{3}{4}} \right) - \left( {\dfrac{1}{4}} \right)$

$ \Rightarrow {\mathbf{sin}}\;{\mathbf{x}} = \left( {\dfrac{1}{2}} \right)$

We know that $\left[ {{\text{sin}}{{30}^ \circ } = \dfrac{1}{2}} \right]$

$ \Rightarrow {\mathbf{sin}}\;{\mathbf{x}} = {\text{sin}}{30^ \circ }$

$ \Rightarrow {\mathbf{x}} = {30^ \circ }$

ii) ${\mathbf{sin}}\;{\mathbf{x}} = {\mathbf{sin}}\;{60^ \circ }.{\mathbf{cos}}\;{30^ \circ } + {\mathbf{cos}}\;{60^ \circ }.{\mathbf{sin}}\;{30^ \circ }$

Ans: ${\mathbf{sin}}\;{\mathbf{x}} = {\mathbf{sin}}\;{60^ \circ }.{\mathbf{cos}}\;{30^ \circ } + {\mathbf{cos}}\;{60^ \circ }.{\mathbf{sin}}\;{30^ \circ }$

$\left[ {{\text{We know that}}:{\mathbf{sin}}\;{{30}^ \circ } = {\mathbf{cos}}\;{{60}^ \circ } = \dfrac{1}{2},\;{\mathbf{sin}}\;{{60}^ \circ } = {\mathbf{cos}}\;{{30}^ \circ } = \dfrac{{\sqrt 3 }}{2}} \right]$

$ \Rightarrow {\mathbf{sin}}\;{\mathbf{x}} = \left( {\dfrac{{\sqrt 3 }}{2}} \right).\left( {\dfrac{{\sqrt 3 }}{2}} \right) + \left( {\dfrac{1}{2}} \right).\left( {\dfrac{1}{2}} \right)$

$ \Rightarrow {\mathbf{sin}}\;{\mathbf{x}} = \left( {\dfrac{3}{4}} \right) + \left( {\dfrac{1}{4}} \right)$

$ \Rightarrow {\mathbf{sin}}\;{\mathbf{x}} = 1$

We know that $\left[ {{\text{sin}}{{90}^ \circ } = 1} \right]$

$ \Rightarrow {\mathbf{sin}}\;{\mathbf{x}} = {\text{sin}}{90^ \circ }$

$ \Rightarrow {\mathbf{x}} = {90^ \circ }$

iii) ${\mathbf{cos}}\;{\mathbf{x}} = {\mathbf{cos}}\;{60^ \circ }.{\mathbf{cos}}\;{30^ \circ } - {\mathbf{sin}}\;{60^ \circ }.{\mathbf{sin}}\;{30^ \circ }$

Ans: ${\mathbf{cos}}\;{\mathbf{x}} = {\mathbf{cos}}\;{60^ \circ }.{\mathbf{cos}}\;{30^ \circ } - {\mathbf{sin}}\;{60^ \circ }.{\mathbf{sin}}\;{30^ \circ }$

$\left[ {{\text{We know that}}:{\mathbf{sin}}\;{{30}^ \circ } = {\mathbf{cos}}\;{{60}^ \circ } = \dfrac{1}{2},\;{\mathbf{sin}}\;{{60}^ \circ } = {\mathbf{cos}}\;{{30}^ \circ } = \dfrac{{\sqrt 3 }}{2}} \right]$

$ \Rightarrow {\mathbf{cos}}\;{\mathbf{x}} = \left( {\dfrac{1}{2}} \right).\left( {\dfrac{{\sqrt 3 }}{2}} \right) - \left( {\dfrac{{\sqrt 3 }}{2}} \right).\left( {\dfrac{1}{2}} \right)$

$ \Rightarrow {\mathbf{cos}}\;{\mathbf{x}} = \left( {\dfrac{{\sqrt 3 }}{4}} \right) - \left( {\dfrac{{\sqrt 3 }}{4}} \right)$

$ \Rightarrow {\mathbf{cos}}\;{\mathbf{x}} = 0$

We know that $\left[ {{\text{cos}}{{90}^ \circ } = 0} \right]$

$ \Rightarrow {\mathbf{cos}}\;{\mathbf{x}} = {\mathbf{co}}{\text{s}}{90^ \circ }$

$ \Rightarrow {\mathbf{x}} = {90^ \circ }$

iv) ${\mathbf{tan}}\;{\mathbf{x}} = \dfrac{{{\mathbf{tan}}\;{{60}^ \circ } - \;{\mathbf{tan}}\;{{30}^ \circ }}}{{1\; + \;{\mathbf{tan}}\;{{60}^ \circ }.\;{\mathbf{tan}}\;{{30}^ \circ }}}$

Ans: ${\mathbf{tan}}\;{\mathbf{x}} = \dfrac{{{\mathbf{tan}}\;{{60}^ \circ } - \;{\mathbf{tan}}\;{{30}^ \circ }}}{{1\; + \;{\mathbf{tan}}\;{{60}^ \circ }.\;{\mathbf{tan}}\;{{30}^ \circ }}}$

$\left[ {{\text{We know that}}:{\mathbf{tan}}\;{{60}^ \circ } = \sqrt 3 \;,\;{\mathbf{tan}}\;{{30}^ \circ } = \dfrac{1}{{\sqrt 3 }}} \right]$

$ \Rightarrow {\mathbf{tan}}\;{\mathbf{x}} = \dfrac{{\sqrt 3 \; - \;\dfrac{1}{{\sqrt 3 }}}}{{1\; + \;\sqrt 3 \;.\;\dfrac{1}{{\sqrt 3 }}}}$

$ \Rightarrow {\mathbf{tan}}\;{\mathbf{x}} = \dfrac{{\dfrac{{3\; - \;1}}{{\sqrt 3 }}}}{{1\; + \;1}}$

$ \Rightarrow {\mathbf{tan}}\;{\mathbf{x}} = \dfrac{{\dfrac{2}{{\sqrt 3 }}}}{2}$

$ \Rightarrow {\mathbf{tan}}\;{\mathbf{x}} = \dfrac{1}{{\sqrt 3 }}$

We know that $\left[ {{\mathbf{tan}}\;{{30}^ \circ } = \dfrac{1}{{\sqrt 3 }}} \right]$

$ \Rightarrow {\mathbf{tan}}\;{\mathbf{x}} = {\mathbf{tan}}\;{30^ \circ }$

$ \Rightarrow {\mathbf{x}} = {30^ \circ }$

v) ${\mathbf{sin}}\;2{\mathbf{x}} = 2{\mathbf{sin}}\;{45^ \circ }.\;{\mathbf{cos}}\;{45^ \circ }$

Ans: ${\mathbf{sin}}\;2{\mathbf{x}} = 2{\mathbf{sin}}\;{45^ \circ }.\;{\mathbf{cos}}\;{45^ \circ }$

$\left[ {{\text{We know that}}:{\mathbf{sin}}\;{{45}^ \circ } = {\mathbf{cos}}\;{{45}^ \circ } = \dfrac{1}{{\sqrt 2 }}} \right]$

$ \Rightarrow {\mathbf{sin}}\;2{\mathbf{x}} = 2 \times \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{{\sqrt 2 }}$

$ \Rightarrow {\mathbf{sin}}\;2{\mathbf{x}} = 2 \times \dfrac{1}{2}$

$ \Rightarrow {\mathbf{sin}}\;2{\mathbf{x}} = 1$

We know that $\left[ {{\text{sin}}{{90}^ \circ } = 1} \right]$

$ \Rightarrow {\mathbf{sin}}\;2{\mathbf{x}} = {\text{sin}}{90^ \circ }$

$ \Rightarrow 2{\mathbf{x}} = {90^ \circ }$

$ \Rightarrow {\mathbf{x}} = {45^ \circ }$

vi) ${\mathbf{sin}}\;3{\mathbf{x}} = 2{\mathbf{sin}}\;{30^ \circ }.\;{\mathbf{cos}}\;{30^ \circ }$

Ans: ${\mathbf{sin}}\;3{\mathbf{x}} = 2{\mathbf{sin}}\;{30^ \circ }.\;{\mathbf{cos}}\;{30^ \circ }$

$\left[ {{\text{We know that}}:{\mathbf{sin}}\;{{30}^ \circ } = \dfrac{1}{2},{\mathbf{cos}}\;{{30}^ \circ } = \dfrac{{\sqrt 3 }}{2}} \right]$

$ \Rightarrow {\mathbf{sin}}\;3{\mathbf{x}} = 2 \times \dfrac{1}{2} \times \dfrac{{\sqrt 3 }}{2}$

$ \Rightarrow {\mathbf{sin}}\;3{\mathbf{x}} = \dfrac{{\sqrt 3 }}{2}$

We know that $\left[ {{\text{sin}}{{60}^ \circ } = \dfrac{{\sqrt 3 }}{2}} \right]$

$ \Rightarrow {\mathbf{sin}}\;3{\mathbf{x}} = {\text{sin}}{60^ \circ }$

$ \Rightarrow 3{\mathbf{x}} = {60^ \circ }$

$ \Rightarrow {\mathbf{x}} = {20^ \circ }$

vii) ${\mathbf{cos}}\;\left( {2{\mathbf{x}} - {6^ \circ }} \right) = {\mathbf{co}}{{\mathbf{s}}^2}{30^ \circ } - {\mathbf{co}}{{\mathbf{s}}^2}{60^ \circ }$

Ans: ${\mathbf{cos}}\;\left( {2{\mathbf{x}} - {6^ \circ }} \right) = {\mathbf{co}}{{\mathbf{s}}^2}{30^ \circ } - {\mathbf{co}}{{\mathbf{s}}^2}{60^ \circ }$

$\left[ {{\text{We know that}}:{\text{co}}{\mathbf{s}}\;{{30}^ \circ } = \dfrac{{\sqrt 3 }}{2},\;{\mathbf{cos}}\;{{60}^ \circ } = \dfrac{1}{2}} \right]$

$ \Rightarrow {\mathbf{cos}}\;\left( {2{\mathbf{x}} - {6^ \circ }} \right) = {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} - {\left( {\dfrac{1}{2}} \right)^2}$

$ \Rightarrow {\mathbf{cos}}\;\left( {2{\mathbf{x}} - {6^ \circ }} \right) = \left( {\dfrac{3}{4}} \right) - \left( {\dfrac{1}{4}} \right)$

$ \Rightarrow {\mathbf{cos}}\;\left( {2{\mathbf{x}} - {6^ \circ }} \right) = \left( {\dfrac{1}{2}} \right)$

We know that $\left[ {{\text{cos}}{{60}^ \circ } = \dfrac{1}{2}} \right]$

$ \Rightarrow {\mathbf{cos}}\;\left( {2{\mathbf{x}} - {6^ \circ }} \right) = {\text{cos}}{60^ \circ }$

$ \Rightarrow 2{\mathbf{x}} - {6^ \circ } = {60^ \circ }$

$ \Rightarrow 2{\mathbf{x}} = {66^ \circ }$

$ \Rightarrow {\mathbf{x}} = {33^ \circ }$

8. In each case, given below, find the value of angle A, where ${0^ \circ } \leqslant {\mathbf{A}} \leqslant {90^ \circ }$.

i) ${\mathbf{sin}}\;\left( {{{90}^ \circ } - 3{\mathbf{A}}} \right).\;{\mathbf{cosec}}\;{42^ \circ } = 1$

Ans: ${\mathbf{sin}}\;\left( {{{90}^ \circ } - 3{\mathbf{A}}} \right).\;{\mathbf{cosec}}\;{42^ \circ } = 1$

[We know that: ${\mathbf{sin}}\;{\mathbf{A}}.\;{\mathbf{cosec}}\;{\mathbf{A}} = 1$]

So, ${\mathbf{sin}}\;\left( {{{90}^ \circ } - 3{\mathbf{A}}} \right) = {\mathbf{sin}}\;{42^ \circ }$

$ \Rightarrow \left( {{{90}^ \circ } - 3{\mathbf{A}}} \right) = {42^ \circ }$

$ \Rightarrow 3{\mathbf{A}} = {90^ \circ } - {42^ \circ }$

$ \Rightarrow 3{\mathbf{A}} = {48^ \circ }$

$ \Rightarrow {\mathbf{A}} = {16^ \circ }$

ii) ${\mathbf{cos}}\;\left( {{{90}^ \circ } - {\mathbf{A}}} \right).\;{\mathbf{sec}}\;{77^ \circ } = 1$

Ans: ${\mathbf{cos}}\;\left( {{{90}^ \circ } - {\mathbf{A}}} \right).\;{\mathbf{sec}}\;{77^ \circ } = 1$

[We know that: ${\mathbf{cos}}\;{\mathbf{A}}.\;{\mathbf{sec}}\;{\mathbf{A}} = 1$]

So, ${\mathbf{cos}}\;\left( {{{90}^ \circ } - {\mathbf{A}}} \right) = {\mathbf{cos}}\;{77^ \circ }$

$ \Rightarrow \left( {{{90}^ \circ } - {\mathbf{A}}} \right) = {77^ \circ }$

$ \Rightarrow {\mathbf{A}} = {90^ \circ } - {77^ \circ }$

$ \Rightarrow {\mathbf{A}} = {13^ \circ }$

9. Prove that: 

i) $\dfrac{{{\mathbf{cos}}\;\left( {{{90}^ \circ }\; - \;{\mathbf{\theta }}} \right){\mathbf{cos}}\;{\mathbf{\theta }}}}{{{\mathbf{cot}}\;{\mathbf{\theta }}}} = 1 - {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{\theta }}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{{\mathbf{cos}}\;\left( {{{90}^ \circ }\; - \;{\mathbf{\theta}}} \right){\mathbf{cos}}\;{\mathbf{\theta }}}}{{{\mathbf{cot}}\;{\mathbf{\theta }}}}$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{\theta }}\;.\;{\mathbf{cos}}\;{\mathbf{\theta }}}}{{\dfrac{{{\mathbf{cos}}\;{\mathbf{\theta }}}}{{{\mathbf{sin}}\;{\mathbf{\theta }}}}}}$                   $\left[ {{\text{We know that}}:{\text{cos}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{sin A}},{\text{cot A}} = \dfrac{{{\text{cos A}}}}{{{\text{sin A}}}}} \right]$

$ = {\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{\theta }}$

$ = 1 - {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{\theta }} = {\mathbf{RHS}}$                                  $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

ii) $\dfrac{{{\mathbf{sin}}\;{\mathbf{\theta }}\;.\;{\mathbf{sin}}\;\left( {{{90}^ \circ }\; - \;{\mathbf{\theta }}} \right)}}{{{\mathbf{cot}}\;\left( {{{90}^ \circ } - \;{\mathbf{\theta }}} \right)}} = 1 - {\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{\theta }}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{\theta }}\;.\;{\mathbf{sin}}\;\left( {{{90}^ \circ }\; - \;{\mathbf{\theta }}} \right)}}{{{\mathbf{cot}}\;\left( {{{90}^ \circ } - \;{\mathbf{\theta }}} \right)}}$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{\theta }}\;.\;{\mathbf{cos}}\;{\mathbf{\theta }}}}{{{\mathbf{tan}}\;{\mathbf{\theta }}}}$           $\left[ {{\text{We know}}:{\text{sin}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{cosA}},{\text{cot}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{tan A}}} \right]$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{\theta }}\;.\;{\mathbf{cos}}\;{\mathbf{\theta }}}}{{\dfrac{{{\mathbf{sin}}\;{\mathbf{\theta }}}}{{{\mathbf{cos}}\;{\mathbf{\theta }}}}}}$                                                            $\left[ {{\text{We know that}}:{\text{tan A}} = \dfrac{{{\text{sin A}}}}{{{\text{cos A}}}}} \right]$

$ = {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{\theta }}$

$ = 1 - {\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{\theta }} = {\mathbf{RHS}}$                                  $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

10. Evaluate: 

$\dfrac{{{\mathbf{sin}}\;{{35}^ \circ }{\mathbf{cos}}\;{{55}^ \circ } + {\mathbf{cos}}\;{{35}^ \circ }{\mathbf{sin}}\;{{55}^ \circ }}}{{{\mathbf{cose}}{{\mathbf{c}}^2}{{10}^ \circ } - \;{\mathbf{ta}}{{\mathbf{n}}^2}{{80}^ \circ }}}$

Ans: $\dfrac{{{\mathbf{sin}}\;{{35}^ \circ }{\mathbf{cos}}\;{{55}^ \circ } + {\mathbf{cos}}\;{{35}^ \circ }{\mathbf{sin}}\;{{55}^ \circ }}}{{{\mathbf{cose}}{{\mathbf{c}}^2}{{10}^ \circ } - \;{\mathbf{ta}}{{\mathbf{n}}^2}{{80}^ \circ }}}$

$ = \dfrac{{{\mathbf{sin}}\;{{\left( {90\; - \;55} \right)}^ \circ }{\mathbf{cos}}\;{{55}^ \circ } + {\mathbf{cos}}\;{{\left( {90\; - 55} \right)}^ \circ }{\mathbf{sin}}\;{{55}^ \circ }}}{{{\mathbf{cose}}{{\mathbf{c}}^2}{{10}^ \circ } - \;{\mathbf{ta}}{{\mathbf{n}}^2}{{\left( {90\; - \;10} \right)}^ \circ }}}$

$\left[ {\because \;{\text{sin}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{cos A}},{\text{cos}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{sin A}},{\text{tan}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{cot A}}} \right]$

$ = \dfrac{{{\mathbf{cos}}\;{{55}^ \circ }.\;{\mathbf{cos}}\;{{55}^ \circ } + \;{\mathbf{sin}}\;{{55}^ \circ }.\;{\mathbf{sin}}\;{{55}^ \circ }}}{{{\mathbf{cose}}{{\mathbf{c}}^2}{{10}^ \circ } - \;{\mathbf{co}}{{\mathbf{t}}^2}{{10}^ \circ }}}$

$ = \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}\;{{55}^ \circ } + \;{\mathbf{si}}{{\mathbf{n}}^2}\;{{55}^ \circ }}}{{{\mathbf{cose}}{{\mathbf{c}}^2}{{10}^ \circ } - \;{\mathbf{co}}{{\mathbf{t}}^2}{{10}^ \circ }}}$

$\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1\;\& \;{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}} - \;{\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{1}{1} = 1$

11. Evaluate: 

${\mathbf{si}}{{\mathbf{n}}^2}{34^ \circ } + {\mathbf{si}}{{\mathbf{n}}^2}{56^ \circ } + 2{\mathbf{tan}}\;{18^ \circ }.{\mathbf{tan}}\;{72^ \circ } - {\mathbf{co}}{{\mathbf{t}}^2}{30^ \circ }$

Ans: ${\mathbf{si}}{{\mathbf{n}}^2}{34^ \circ } + {\mathbf{si}}{{\mathbf{n}}^2}{56^ \circ } + 2{\mathbf{tan}}\;{18^ \circ }.{\mathbf{tan}}\;{72^ \circ } - {\mathbf{co}}{{\mathbf{t}}^2}{30^ \circ }$

$ = {\mathbf{si}}{{\mathbf{n}}^2}{\left( {90 - 56} \right)^ \circ } + {\mathbf{si}}{{\mathbf{n}}^2}{56^ \circ } + 2{\mathbf{tan}}\;{18^ \circ }.{\mathbf{tan}}\;{\left( {90 - 18} \right)^ \circ } - {\mathbf{co}}{{\mathbf{t}}^2}{30^ \circ }$

$\left[ {\because \;{\text{sin}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{cos A}},{\text{tan}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{cot A}},{\text{cot}}{{30}^ \circ } = \sqrt 3 {\text{}}} \right]$

$ = {\mathbf{co}}{{\mathbf{s}}^2}{56^ \circ } + {\mathbf{si}}{{\mathbf{n}}^2}{56^ \circ } + 2{\mathbf{tan}}\;{18^ \circ }.{\mathbf{cot}}\;{18^ \circ } - {\left( {\sqrt 3 } \right)^2}$

$\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1\;\& \;{\mathbf{tan}}\;{\mathbf{A}}.{\mathbf{cot}}\;{\mathbf{A}} = 1} \right]$

$ = 1 + \left( {2 \times 1} \right) - 3$

$ = 0$

12. Evaluate: 

${\mathbf{cose}}{{\mathbf{c}}^2}{57^ \circ } - {\mathbf{ta}}{{\mathbf{n}}^2}{33^ \circ } + {\mathbf{cos}}\;{44^ \circ }.{\mathbf{cosec}}\;{46^ \circ } - \sqrt 2 \;{\mathbf{cos}}\;{45^ \circ } - {\mathbf{ta}}{{\mathbf{n}}^2}{60^ \circ }$

Ans: ${\mathbf{cose}}{{\mathbf{c}}^2}{57^ \circ } - {\mathbf{ta}}{{\mathbf{n}}^2}{33^ \circ } + {\mathbf{cos}}\;{44^ \circ }.{\mathbf{cosec}}\;{46^ \circ } - \sqrt 2 \;{\mathbf{cos}}\;{45^ \circ } - {\mathbf{ta}}{{\mathbf{n}}^2}{60^ \circ }$

$\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:\;{\text{cos}}{{45}^ \circ } = \dfrac{1}{{\sqrt 2 }}\;,{\text{tan}}{{60}^ \circ } = \sqrt 3 {\text{}}} \right]$

$ = {\mathbf{cose}}{{\mathbf{c}}^2}{57^ \circ } - {\mathbf{ta}}{{\mathbf{n}}^2}{\left( {90 - 57} \right)^ \circ } + {\mathbf{cos}}\;{44^ \circ }.{\mathbf{cosec}}\;{\left( {90 - 44} \right)^ \circ } - \sqrt 2  \times \dfrac{1}{{\sqrt 2 }} - {\left( {\sqrt 3 } \right)^2}$

$\left[ {\because \;{\mathbf{cosec}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{sec A}},{\text{tan}}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{cot A}}} \right]$

$ = {\mathbf{cose}}{{\mathbf{c}}^2}{57^ \circ } - {\mathbf{co}}{{\mathbf{t}}^2}{57^ \circ } + {\mathbf{cos}}\;{44^ \circ }.{\mathbf{sec}}\;{44^ \circ } - 1 - 3$

$\left[{{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}} - {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} = 1\;\& \;{\mathbf{cos}}\;{\mathbf{A}}.{\mathbf{sec}}\;{\mathbf{A}} = 1} \right]$

$ = 1 + 1 - 1 - 3$

$ =  - 2$

Exercise - 21(D)

1. Use table to find sine of:

i) ${21^ \circ }$

Ans: ${\mathbf{sin}}\;{21^ \circ }$

According to the natural sine table,

${\text{sin}}{21^ \circ } = 0.3584$

ii) ${34^ \circ }42'$

Ans: ${\mathbf{sin}}\;{34^ \circ }42'$

According to the natural sine table,

${\text{sin}}{34^ \circ }42{\text{'}} = 0.5693$

iii) ${47^ \circ }32'$

Ans: ${\mathbf{sin}}\;{47^ \circ }32'$

According to the natural sine table,

${\text{sin}}{47^ \circ }32{\text{'}} = {\text{sin}}({47^ \circ }30{\text{'}} + 2{\text{'}})$

$ \Rightarrow {\text{sin}}{47^ \circ }32{\text{'}} = 0.7373 + 0.0004$

$ \Rightarrow {\text{sin}}{47^ \circ }32{\text{'}} = 0.7377$

iv) ${62^ \circ }57'$

Ans: ${\mathbf{sin}}\;{62^ \circ }57'$

According to the natural sine table,

${\text{sin}}{62^ \circ }57{\text{'}} = {\text{sin}}({62^ \circ }54{\text{'}} + 3{\text{'}})$

$ \Rightarrow {\text{sin}}{62^ \circ }57{\text{'}} = 0.8902 + 0.0004$

$ \Rightarrow {\text{sin}}{62^ \circ }57{\text{'}} = 0.8906$

v) ${10^ \circ }20' + {20^ \circ }45'$

Ans: ${\mathbf{sin}}\;({10^ \circ }20' + {20^ \circ }45')$

According to the natural sine table,

${\text{sin}}({10^ \circ }20{\text{'}} + {20^ \circ }45{\text{'}}) = {\text{sin}}({30^ \circ }65{\text{'}})$

$ \Rightarrow {\text{sin}}({10^ \circ }20{\text{'}} + {20^ \circ }45{\text{'}}) = {\text{sin}}({31^ \circ }5{\text{'}})$

$ \Rightarrow {\text{sin}}({10^ \circ }20{\text{'}} + {20^ \circ }45{\text{'}}) = 0.5150 + 0.0012$

$ \Rightarrow {\text{sin}}({10^ \circ }20{\text{'}} + {20^ \circ }45{\text{'}}) = 0.5162$

2. Use table to find cosine of:

i) ${2^ \circ }4'$

Ans: ${\mathbf{cos}}\;{2^ \circ }4'$

According to the natural cosine table,

${\text{cos}}{2^ \circ }4{\text{'}} = 0.9994 - 0.0001$

$ \Rightarrow {\text{cos}}{2^ \circ }4{\text{'}} = 0.9993$

ii) ${8^ \circ }12'$

Ans: ${\mathbf{cos}}\;{8^ \circ }12'$

According to the natural cosine table,

${\text{cos}}{8^ \circ }12{\text{'}} = 0.9898$

iii) ${26^ \circ }32'$

Ans: ${\mathbf{cos}}\;{26^ \circ }32'$

According to the natural cosine table,

${\text{cos}}{26^ \circ }32{\text{'}} = {\text{cos}}({26^ \circ }30{\text{'}} + 2{\text{'}})$

$ \Rightarrow {\text{cos}}{26^ \circ }32{\text{'}} = 0.8949 - 0.0003$

$ \Rightarrow {\text{cos}}{26^ \circ }32{\text{'}} = 0.8946$

iv) ${65^ \circ }41'$

Ans: ${\mathbf{cos}}\;{65^ \circ }41'$

According to the natural cosine table,

${\text{cos}}{65^ \circ }41{\text{'}} = {\text{cos}}({65^ \circ }36{\text{'}} + 5{\text{'}})$

$ \Rightarrow {\text{cos}}{65^ \circ }41{\text{'}} = 0.4131 - 0.0013$

$ \Rightarrow {\text{cos}}{65^ \circ }41{\text{'}} = 0.4118$

v) ${9^ \circ }23' + {15^ \circ }54'$

Ans: ${\mathbf{cos}}\;({9^ \circ }23' + {15^ \circ }54')$

According to the natural cosine table,

${\text{cos}}\left( {{9^ \circ }23{\text{'}} + {{15}^ \circ }54{\text{'}}} \right) = {\text{cos}}({24^ \circ }77{\text{'}})$

$ \Rightarrow {\text{cos}}\left( {{9^ \circ }23{\text{'}} + {{15}^ \circ }54{\text{'}}} \right) = {\text{cos}}({25^ \circ }17{\text{'}})$

$ \Rightarrow {\text{cos}}\left( {{9^ \circ }23{\text{'}} + {{15}^ \circ }54{\text{'}}} \right) = {\text{cos}}({25^ \circ }12{\text{'}} + 5{\text{'}})$

$ \Rightarrow {\text{cos}}\left( {{9^ \circ }23{\text{'}} + {{15}^ \circ }54{\text{'}}} \right) = 0.9048 - 0.0006$

$ \Rightarrow {\text{cos}}\left( {{9^ \circ }23{\text{'}} + {{15}^ \circ }54{\text{'}}} \right) = 0.9042$

3. Use trigonometrical tables to find tangent of:

i) ${37^ \circ }$

Ans: ${\mathbf{tan}}\;{37^ \circ }$

According to the natural tangent table,

${\text{tan}}{37^ \circ } = 0.7536$

ii) ${42^ \circ }18'$

Ans: ${\mathbf{tan}}\;{42^ \circ }18'$

According to the natural tangent table,

${\text{tan}}{42^ \circ }18{\text{'}} = 0.9099$

iii) ${17^ \circ }27'$

Ans: ${\mathbf{tan}}\;{17^ \circ }27'$

According to the natural tangent table,

${\text{tan}}{17^ \circ }27{\text{'}} = {\text{tan}}({17^ \circ }24{\text{'}} + 3{\text{'}})$

$ \Rightarrow {\text{tan}}{17^ \circ }27{\text{'}} = 0.3134 + 0.0010$

$ \Rightarrow {\text{tan}}{17^ \circ }27{\text{'}} = 0.3144$

4. Use tables to find the acute angle ${{\theta}}$, if the value of ${\mathbf{sin}}\;{\mathbf{\theta }}$ is:

i) 0.4848

Ans: ${\mathbf{sin}}\;{\mathbf{\theta }} = 0.4848$

From the sine table, it is clear that ${\text{sin}}{29^ \circ } = 0.4848$

${{\sin \theta }} = {\text{sin}}{29^ \circ }$

Hence, ${{\theta }} = {29^ \circ }$

ii) 0.3827

Ans: ${\mathbf{sin}}\;{\mathbf{\theta }} = 0.3827$

From the sine table, it is clear that ${\text{sin}}{22^ \circ }30{\text{'}} = 0.3827$

${{\sin\theta }} = {\text{sin}}{22^ \circ }30{\text{'}}$

Hence, ${{\theta }} = {22^ \circ }30{\text{'}}$

iii) 0.6525

Ans: ${\mathbf{sin}}\;{\mathbf{\theta }} = 0.6525$

From the sine table, it is clear that ${\text{sin}}{40^ \circ }42{\text{'}} = 0.6521$

${\mathbf{sin}}\;{\mathbf{\theta }} - {\mathbf{sin}}\;{40^ \circ }42' = 0.6525 - 0.6521$

So, the mean difference is $0.0004$.

From the table, difference of $2{\text{'}} = 0.0004$

Hence, ${{\theta }} = {40^ \circ }42' + 2' = {40^ \circ }44'$

5. Use tables to find the acute angle ${{\theta}}$, if the value of ${\mathbf{cos}}\;{\mathbf{\theta }}$ is:

i) 0.9848

Ans: ${\mathbf{cos}}\;{\mathbf{\theta }} = 0.9848$

From the cosine table, it is clear that ${\text{cos}}{10^ \circ } = 0.9848$

${{cos\theta }} = {\text{cos}}{10^ \circ }$

Hence, ${{\theta}} = {10^ \circ }$

ii) 0.9574

Ans: ${\mathbf{cos}}\;{\mathbf{\theta }} = 0.9574$

From the cosine table, it is clear that ${\text{cos}}{16^ \circ }48{\text{'}} = 0.9573$

${\mathbf{cos}}\;{\mathbf{\theta }} - {\mathbf{cos}}\;{16^ \circ }48{\text{'}} = 0.9574 - 0.9573$

So, the mean difference is $0.0001$.

From the table, difference of $1{\text{'}} = 0.0001$

Hence, ${{\theta}} = {16^ \circ }48{\text{'}} - 1' = {16^ \circ }47'$

iii) 0.6885

Ans: ${\mathbf{cos}}\;{\mathbf{\theta }} = 0.6885$

From the cosine table, it is clear that ${\text{cos}}{46^ \circ }30{\text{'}} = 0.6884$

${\mathbf{cos}}\;{\mathbf{\theta }} - {\mathbf{cos}}\;{46^ \circ }30{\text{'}} = 0.6885 - 0.6884$

So, the mean difference is $0.0001$.

From the table, difference of $1{\text{'}} = 0.0001$

Hence, ${{\theta }} = {46^ \circ }30{\text{'}} - 1' = {46^ \circ }29'$

6. Use tables to find the acute angle ${{\theta}}$, if the value of ${\mathbf{tan}}\;{\mathbf{\theta }}$ is:

i) 0.2419

Ans: ${\mathbf{tan}}\;{\mathbf{\theta }} = 0.2419$

From the tangent table, it is clear that ${\text{tan}}{13^ \circ }36{\text{'}} = 0.2419$

Hence, ${{\theta }} = {13^ \circ }36{\text{'}}$

ii) 0.4741

Ans: ${\mathbf{tan}}\;{\mathbf{\theta }} = 0.4741$

From the tangent table, it is clear that ${\text{tan}}\;{25^ \circ }18{\text{'}} = 0.4727$

${\mathbf{tan}}\;{\mathbf{\theta }} - {\mathbf{tan}}\;{25^ \circ }18' = 0.4741 - 0.4727$

So, the mean difference is $0.0014$.

From the table, difference of $4{\text{'}} = 0.0014$

Hence, ${{\theta}} = {25^ \circ }18{\text{'}} + 4' = {25^ \circ }22'$

iii) 0.7391

Ans: ${\mathbf{tan}}\;{\mathbf{\theta }} = 0.7391$

From the tangent table, it is clear that ${\text{tan}}\;{36^ \circ }24{\text{'}} = 0.7373$

${\mathbf{tan}}\;{\mathbf{\theta }} - {\mathbf{tan}}\;{36^ \circ }24{\text{'}} = 0.7391 - 0.7391$

So, the mean difference is $0.0018$.

From the table, difference of $4{\text{'}} = 0.0018$

Hence, ${{\theta }} = {36^ \circ }24{\text{'}} + 4' = {36^ \circ }28{\text{'}}$

Exercise - 21(E)

1. Prove the following identities:

i) $\dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}}} + \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{A}}}} = \dfrac{{2{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; - \;1}}$

Ans: Taking LHS -

${\text{LHS}} = \dfrac{1}{{{\text{cos A}} + {\text{sin A}}}} + \dfrac{1}{{{\text{cos A}} - {\text{sin A}}}}$

$ = \dfrac{{{\text{cos A}} - {\text{sin A}} + {\text{cos A}} + {\text{sin A}}}}{{\left( {{\text{cos A}} + {\text{sin A}}} \right)\left( {{\text{cos A}} - {\text{sin A}}} \right)}}$

$ = \dfrac{{2{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; - \;1}} = {\mathbf{RHS}}$                                                  $\left[ {\because {\text{}}{{\text{a}}^2} - {{\text{b}}^2} = \left( {{\text{a}} + {\text{b}}} \right)\left( {{\text{a}} - {\text{b}}} \right)} \right]$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

ii) ${\mathbf{cosec}}\;{\mathbf{A}} - {\mathbf{cot}}\;{\mathbf{A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}$

Ans: Taking LHS -

${\text{LHS}} = {\text{cosec A}} - {\text{cot A}}$

$ = \dfrac{1}{{{\text{sin A}}}} - \dfrac{{{\text{cos A}}}}{{{\text{sinA}}}}$                                               $\left[ {\because {\text{cosec A}} = \dfrac{1}{{{\text{sin A}}}},{\text{cot A}} = \dfrac{{{\text{cos A}}}}{{{\text{sin A}}}}} \right]$

$ = \dfrac{{1{\text{}} - {\text{cos A}}}}{{{\text{sin A}}}}$

$ = \dfrac{{1{\text{}} - {\text{cos A}}}}{{{\text{sin A}}}} \times \left( {\dfrac{{1{\text{}} + {\text{cos A}}}}{{1{\text{}} + {\text{cos A}}}}} \right)$

$ = \dfrac{{1{\text{}} - {\text{}}{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\text{sin A}}\left( {1{\text{}} + {\text{cos A}}} \right)}}$                                                       $\left[ {\because {\text{}}{{\text{a}}^2} - {{\text{b}}^2} = \left( {{\text{a}} + {\text{b}}} \right)\left( {{\text{a}} - {\text{b}}} \right)} \right]$

$ = \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\text{sin A}}\left( {1{\text{}} + {\text{cos A}}} \right)}}$                                            $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$         

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}} = {\mathbf{RHS}}$                                              

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

iii) $1 - \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}} = {\mathbf{cos}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\text{LHS}} = 1 - \dfrac{{{\text{si}}{{\text{n}}^2}{\text{A}}}}{{1{\text{}} + {\text{cosA}}}}$

$ = 1 - \dfrac{{1{\text{}} - {\text{}}{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{1{\text{}} + {\text{cosA}}}}$                                            $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = 1 - \dfrac{{\left( {1{\text{}} - {\text{cos A}}} \right)\left( {1{\text{}} + {\text{cos A}}} \right)}}{{1{\text{}} + {\text{cos A}}}}$                                       $\left[ {\because {\text{}}{{\text{a}}^2} - {{\text{b}}^2} = \left( {{\text{a}} + {\text{b}}} \right)\left( {{\text{a}} - {\text{b}}} \right)} \right]$

$ = 1 - \left( {1{\text{}} - {\text{cos A}}} \right)$

$ = {\mathbf{cos}}\;{\mathbf{A}} = {\mathbf{RHS}}$                                              

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

iv) $\dfrac{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}} = 2{\mathbf{cosec}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\text{LHS}} = \dfrac{{1{\text{}} - {\text{cos A}}}}{{{\text{sin A}}}} + \dfrac{{{\text{sin A}}}}{{1{\text{}} - {\text{cos A}}}}$

$ = \dfrac{{{{\left( {1{\text{}} - {\text{ cos A}}} \right)}^2}{\text{}} + {\text{si}}{{\text{n}}^2}{\text{A}}}}{{{\text{sin A}}\left( {1{\text{}} - {\text{cos A}}} \right)}}$                                            

$ = \dfrac{{1{\text{}} + {\text{co}}{{\text{s}}^2}{\text{A}} - {\text{}}2{\text{cos A}} + {\text{si}}{{\text{n}}^2}{\text{A}}}}{{{\text{sin A}}\left( {1{\text{}} - {\text{cos A}}} \right)}}$                                         $\left[ {\because {\text{}}{{\text{a}}^2} + {{\text{b}}^2} + 2{\text{ab}} = {{({\text{a}} + {\text{b}})}^2}} \right]$

$ = \dfrac{{1{\text{}} + {\text{}}1{\text{}} - {\text{}}2{\text{cos A}}}}{{{\text{sin A}}\left( {1{\text{}} - {\text{cos A}}} \right)}}$                                             $\left[ {{\mathbf{We}}\;{\mathbf{know}}\;{\mathbf{that}}:{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{{2\left( {1{\text{}} - {\text{cos A}}} \right)}}{{{\text{sin A}}\left( {1{\text{}} - {\text{cos A}}} \right)}}$

$ = 2{\mathbf{cosec}}\;{\mathbf{A}} = {\mathbf{RHS}}$                                                                   $\left[ {\because {\text{sin A}} = \dfrac{1}{{{\text{cosec A}}}}} \right]$                                

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

v) $\dfrac{{{\mathbf{cot}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{tan}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{tan}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{cot}}\;{\mathbf{A}}}} = 1 + {\mathbf{tan}}\;{\mathbf{A}} + {\mathbf{cot}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\text{LHS}} = \dfrac{{{\text{cot A}}}}{{1{\text{}} - {\text{tan A}}}} + \dfrac{{{\text{tan A}}}}{{1{\text{}} - {\text{cot A}}}}$

$ = \dfrac{{\dfrac{1}{{{\text{tan A}}}}}}{{1{\text{}} - {\text{ tan A}}}} + \dfrac{{{\text{tan A}}}}{{1{\text{}} - {\text{}}\dfrac{1}{{{\text{tan A}}}}}}$                                                              $\left[ {\because {\text{tan A}} = \dfrac{1}{{{\text{cot A}}}}} \right]$     

$ = \dfrac{1}{{{\text{tan A}}\left( {1{\text{}} - {\text{tan A}}} \right)}} + \dfrac{{{\text{tan A}}}}{{\dfrac{{{\text{tan A}} - {\text{}}1}}{{{\text{tan A}}}}}}$                                         

$ = \dfrac{1}{{{\text{tan A}}\left( {1{\text{}} - {\text{tan A}}} \right)}} - \dfrac{{{\text{ta}}{{\text{n}}^2}{\text{A}}}}{{1{\text{}} - {\text{tan A}}}}$                                             

$ = \dfrac{{1{\text{}} - {\text{ta}}{{\text{n}}^3}{\text{A}}}}{{{\text{tan A}}\left( {1{\text{}} - {\text{tan A}}} \right)}}$

$ = \dfrac{{\left( {1{\text{}} - {\text{tan A}}} \right)\left( {1{\text{}} + {\text{ta}}{{\text{n}}^2}{\text{A}} + {\text{tan A}}} \right)}}{{{\text{tan\;A\;}}\left( {1{\text{}} - {\text{ tan A}}} \right)}}$                       $\left[ {\because {\text{}}{{\text{a}}^3} - {{\text{b}}^3} = \left( {{\text{a}} - {\text{b}}} \right)\left( {{{\text{a}}^2} + {{\text{b}}^2} + {\text{ab}}} \right)} \right]$

$ = \dfrac{{\left( {1{\text{}} + {\text{ta}}{{\text{n}}^2}{\text{A}} + {\text{tanA}}} \right)}}{{{\text{tanA}}}}$

$ = {\mathbf{cot}}\;{\mathbf{A}} + {\mathbf{tan}}\;{\mathbf{A}} + 1 = {\mathbf{RHS}}$                                                 $\left[ {\because {\text{cot A}} = \dfrac{1}{{{\text{tan A}}}}} \right]$                                

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

vi) $\dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}} + {\mathbf{tan}}\;{\mathbf{A}} = {\mathbf{sec}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\text{LHS}} = \dfrac{{{\text{cos A}}}}{{1{\text{}} + {\text{sin A}}}} + {\text{tan A}}$

$ = \dfrac{{{\text{cos A}}}}{{1{\text{}} + {\text{sinA}}}} + \dfrac{{{\text{sin A}}}}{{{\text{cosA}}}}$                                                                $\left[ {\because {\text{tan A}} = \dfrac{{{\text{sin A}}}}{{{\text{cos A}}}}} \right]$  

$ = \dfrac{{{\text{co}}{{\text{s}}^2}{\text{A}} + {\text{ sin A}} + {\text{si}}{{\text{n}}^2}{\text{A}}}}{{\left( {1{\text{}} + {\text{sinA}}} \right){\text{cos A}}}}$                                         

$ = \dfrac{{1{\text{}} + {\text{sin A}}}}{{\left( {1{\text{}} + {\text{sin A}}} \right){\text{cos A}}}}$                                                            $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{1}{{{\text{cos A}}}}$

$ = {\mathbf{sec}}\;{\mathbf{A}} = {\mathbf{RHS}}$                                                                   $\left[ {\because {\text{sec A}} = \dfrac{1}{{{\text{cos A}}}}} \right]$                                

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

vii) $\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}} - {\mathbf{cot}}\;{\mathbf{A}} = {\mathbf{cosec}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\text{LHS}} = \dfrac{{{\text{sin A}}}}{{1{\text{}} - {\text{cos A}}}} - {\text{cot A}}$

$ = \dfrac{{{\text{sin A}}}}{{1{\text{}} - {\text{cos A}}}} - \dfrac{{{\text{cos A}}}}{{{\text{sin A}}}}$                                                                $\left[ {\because {\text{cot A}} = \dfrac{{{\text{cos A}}}}{{{\text{sin A}}}}} \right]$  

$ = \dfrac{{{\text{si}}{{\text{n}}^2}{\text{A}} - {\text{cosA}} + {\text{co}}{{\text{s}}^2}{\text{A}}}}{{\left( {1{\text{}} - {\text{cos A}}} \right){\text{sinA}}}}$                                         

$ = \dfrac{{1{\text{}} - {\text{cos A}}}}{{\left( {1{\text{}} - {\text{cos A}}} \right){\text{sin A}}}}$                                                            $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{1}{{{\text{sin A}}}}$

$ = {\text{co}}{\mathbf{sec}}\;{\mathbf{A}} = {\mathbf{RHS}}$                                                                   $\left[ {\because {\text{cosec A}} = \dfrac{1}{{{\text{sin A}}}}} \right]$                                

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

viii) $\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{cos}}\;{\mathbf{A}}\; + \;1}}{{{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{A}}\; - \;1}} = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}$

Ans: Taking LHS -

${\text{LHS}} = \dfrac{{{\text{sin A}} - {\text{cos A}} + {\text{}}1}}{{{\text{sin A}} + {\text{cos A}} - {\text{}}1}}$

$ = \dfrac{{{\text{sin A}} - {\text{ cos A}} + {\text{}}1}}{{{\text{sin A}} + {\text{cos A}} - {\text{}}1}} \times \dfrac{{{\text{sin A}} - {\text{}}\left( {{\text{cos A}} - {\text{}}1} \right)}}{{{\text{sin A}} - {\text{}}\left( {{\text{cos A}} - {\text{}}1} \right)}}$                                                                

$ = \dfrac{{{{\left( {{\text{sin A}} - {\text{ cos A}} + {\text{}}1} \right)}^2}}}{{{\text{si}}{{\text{n}}^2}{\text{A}} - {\text{}}{{\left( {{\text{cos A}} - {\text{}}1} \right)}^2}}}$                                                  $\left[ {\because {\text{}}{{\text{a}}^2} - {{\text{b}}^2} = \left( {{\text{a}} + {\text{b}}} \right)\left( {{\text{a}} - {\text{b}}} \right)} \right]$

$ = \dfrac{{{\text{si}}{{\text{n}}^2}{\text{A}} + {\text{}}{{({\text{cos A}} - {\text{}}1)}^2}{\text{}} - {\text{}}2{\text{sin A}}\left( {{\text{cos A}} - {\text{}}1} \right)}}{{{\text{si}}{{\text{n}}^2}{\text{A}} - {\text{co}}{{\text{s}}^2}{\text{A}} - {\text{}}1{\text{}} + {\text{}}2{\text{cos A}}}}$                         $\left[ {\because {\text{}}{{\text{a}}^2} + {{\text{b}}^2} - 2{\text{ab}} = {{({\text{a}} - {\text{b}})}^2}} \right]$                     

$ = \dfrac{{{\text{si}}{{\text{n}}^2}{\text{A}} + {\text{co}}{{\text{s}}^2}{\text{A}} + {\text{}}1{\text{}} - {\text{}}2{\text{cos A}} - {\text{}}2{\text{sin A cos A}} + {\text{}}2{\text{sin A}}}}{{{\text{si}}{{\text{n}}^2}{\text{A}} - {\text{co}}{{\text{s}}^2}{\text{A}} - {\text{}}1{\text{}} + {\text{}}2{\text{cos A}}}}$            $\left[ {\because {\text{}}{{\text{a}}^2} + {{\text{b}}^2} - 2{\text{ab}} = {{({\text{a}} - {\text{b}})}^2}} \right]$

$ = \dfrac{{1{\text{}} + {\text{}}1{\text{}} - {\text{}}2{\text{cos A}} - {\text{}}2{\text{sin A cos A}} + {\text{}}2{\text{sin A}}}}{{1{\text{}} - {\text{co}}{{\text{s}}^2}{\text{A}} - {\text{co}}{{\text{s}}^2}{\text{A}} - {\text{}}1{\text{}} + {\text{}}2{\text{cos A}}}}$                                        $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{{2\left( {1{\text{}} - {\text{cos A}}} \right){\text{}} + {\text{}}2{\text{sin A}}\left( { - {\text{cos A}} + {\text{}}1} \right)}}{{ - {\text{}}2{\text{co}}{{\text{s}}^2}{\text{A}} + {\text{}}2{\text{cos A}}}}$

$ = \dfrac{{2\left( {1{\text{}} - {\text{cos A}}} \right){\text{}}\left( {1{\text{}} + {\text{sin A}}} \right)}}{{2{\text{cos A}}\left( {1{\text{}} - {\text{cos A}}} \right)}}$

$ = \dfrac{{\left( {1{\text{}} + {\text{sin A}}} \right)}}{{{\text{cos A}}}}$

$ = \dfrac{{\left( {1{\text{}} + {\text{sin A}}} \right)}}{{{\text{cos A}}}} \times \dfrac{{\left( {1{\text{}} - {\text{sin A}}} \right)}}{{\left( {1{\text{}} - {\text{sin A}}} \right)}}$

$ = \dfrac{{\left( {1{\text{}} - {\text{si}}{{\text{n}}^2}{\text{A}}} \right)}}{{{\text{cos A}}\left( {1{\text{}} - {\text{sin A}}} \right)}}$                                                       $\left[ {\because {\text{}}{{\text{a}}^2} - {{\text{b}}^2} = \left( {{\text{a}} + {\text{b}}} \right)\left( {{\text{a}} - {\text{b}}} \right)} \right]$

$ = \dfrac{{{\text{co}}{{\text{s}}^2}{\text{A}}}}{{{\text{cos A}}\left( {1{\text{}} - {\text{ sin A}}} \right)}}$                                                                    $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}} = {\mathbf{RHS}}$                                                                                             

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

ix) $\sqrt {\dfrac{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}}  = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}$

Ans: Taking LHS -

${\text{LHS}} = \sqrt {\dfrac{{1{\text{}} + {\text{sin A}}}}{{1{\text{}} - {\text{sin A}}}}} $

$ = \sqrt {\dfrac{{1{\text{}} + {\text{sin A}}}}{{1{\text{}} - {\text{sin A}}}} \times \dfrac{{1{\text{}} - {\text{sin A}}}}{{1{\text{}} - {\text{sin A}}}}} $                                                                

$ = \sqrt {\dfrac{{{1^2}{\text{}} - {\text{si}}{{\text{n}}^2}{\text{A}}}}{{{{(1{\text{}} - {\text{ sin A}})}^2}}}} $                                                  $\left[ {\because {\text{}}{{\text{a}}^2} - {{\text{b}}^2} = \left( {{\text{a}} + {\text{b}}} \right)\left( {{\text{a}} - {\text{b}}} \right)} \right]$

$ = \sqrt {\dfrac{{{\text{co}}{{\text{s}}^2}{\text{A}}}}{{{{(1{\text{}} - {\text{sin A}})}^2}}}} $                                                               $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$                

$ = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}} = {\mathbf{RHS}}$                                                                                             

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

x) $\sqrt {\dfrac{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}}  = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}$

Ans: Taking LHS -

${\text{LHS}} = \sqrt {\dfrac{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}} $

$ = \sqrt {\dfrac{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}} \times \dfrac{{1{\text{}} + {\text{cos A}}}}{{1{\text{}} + {\text{cos A}}}}} $                                                                

$ = \sqrt {\dfrac{{{1^2}{\text{}} - {\text{\;co}}{{\text{s}}^2}{\text{A}}}}{{{{(1{\text{}} + {\text{cos A}})}^2}}}} $                                                  $\left[ {\because {\text{}}{{\text{a}}^2} - {{\text{b}}^2} = \left( {{\text{a}} + {\text{b}}} \right)\left( {{\text{a}} - {\text{b}}} \right)} \right]$

$ = \sqrt {\dfrac{{{\text{si}}{{\text{n}}^2}{\text{A}}}}{{{{(1{\text{}} + {\text{cos A}})}^2}}}} $                                                               $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$                

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{cos}}\;{\mathbf{A}}}} = {\mathbf{RHS}}$                                                                                             

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

xi) $\dfrac{{1\; + \;{{\left( {{\mathbf{sec}}\;{\mathbf{A}}\; - \;{\mathbf{tan}}\;{\mathbf{A}}} \right)}^2}}}{{{\mathbf{cosec}}\;{\mathbf{A}}\left( {{\mathbf{sec}}\;{\mathbf{A}}\; - \;{\mathbf{tan}}\;{\mathbf{A}}} \right)}} = 2{\mathbf{tan}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\text{LHS}} = \dfrac{{1{\text{}} + {\text{}}{{\left( {{\text{sec A}} - {\text{tan A}}} \right)}^2}}}{{{\text{cosec A}}\left( {{\text{sec A}} - {\text{tan A}}} \right)}}$

$ = \dfrac{{{\text{se}}{{\text{c}}^2}{\text{A}} - {\text{ta}}{{\text{n}}^2}{\text{A}} + {\text{}}{{\left( {{\text{sec A}} - {\text{tan A}}} \right)}^2}}}{{{\text{cosec A}}\left( {{\text{sec A}} - {\text{tan A}}} \right)}}$                                             $\left[ {\because {\text{se}}{{\text{c}}^2}{\text{A}} - {\text{ta}}{{\text{n}}^2}{\text{A}} = 1} \right]$                                        

$ = \dfrac{{\left( {{\text{sec A}} - {\text{tan A}}} \right){\text{}}\left( {{\text{sec A}} + {\text{tan A}}} \right){\text{}} + {\text{}}{{\left( {{\text{sec A}} - {\text{tan A}}} \right)}^2}}}{{{\text{cosec A}}\left( {{\text{sec A}} - {\text{tan A}}} \right)}}$               $\left[ {\because {\text{}}{{\text{a}}^2} - {{\text{b}}^2} = \left( {{\text{a}} + {\text{b}}} \right)\left( {{\text{a}} - {\text{b}}} \right)} \right]$

$ = \dfrac{{\left( {{\text{sec A}} - {\text{tan A}}} \right){\text{}}\left[ {\left( {{\text{sec A}} + {\text{tan A}}} \right){\text{}} + {\text{}}\left( {{\text{sec A}} - {\text{tan A}}} \right)} \right]}}{{{\text{cosec A}}\left( {{\text{sec A}} - {\text{tan A}}} \right)}}$

$ = \dfrac{{\left[ {\left( {{\text{sec A}} + {\text{ tan A}}} \right){\text{}} + {\text{}}\left( {{\text{sec A}} - {\text{tan A}}} \right)} \right]}}{{{\text{cosec A}}}}$

$ = \dfrac{{2{\text{sec A}}}}{{{\text{cosec A}}}}$

$ = \dfrac{{2{\text{sin A}}}}{{{\text{cos A}}}}$                                                             $\left[ {\because {\text{ cosec A}} = \dfrac{1}{{{\text{sin A}}}},{\text{ sec A}} = \dfrac{1}{{{\text{cos A}}}}} \right]$

$ = 2{\mathbf{tan}}\;{\mathbf{A}} = {\mathbf{RHS}}$                                                                         $\left[ {\because {\text{ tan A}} = \dfrac{{{\text{sin A}}}}{{{\text{cos A}}}}} \right]$                                                     

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

xii) $\dfrac{{{{\left( {{\mathbf{cosec}}\;{\mathbf{A}}\; - \;{\mathbf{cot}}\;{\mathbf{A}}} \right)}^2}\; + \;1}}{{{\mathbf{sec}}\;{\mathbf{A}}\left( {{\mathbf{cosec}}\;{\mathbf{A}}\; - \;{\mathbf{cot}}\;{\mathbf{A}}} \right)}} = 2{\mathbf{cot}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\text{LHS}} = \dfrac{{{{\left( {{\text{cosec A}} - {\text{cot A}}} \right)}^2}{\text{}} + {\text{}}1}}{{{\text{sec A}}\left( {{\text{cosec A}} - {\text{cot A}}} \right)}}$

$ = \dfrac{{{{\left( {{\text{cosec A}} - {\text{cot A}}} \right)}^2}{\text{}} + {\text{cose}}{{\text{c}}^2}{\text{A}} - {\text{co}}{{\text{t}}^2}{\text{A}}}}{{{\text{sec A}}\left( {{\text{cosec A}} - {\text{cot A}}} \right)}}$                                     $\left[ {\because {\text{cose}}{{\text{c}}^2}{\text{A}} - {\text{co}}{{\text{t}}^2}{\text{A}} = 1} \right]$                                        

$ = \dfrac{{{{\left( {{\text{cosec A}} - {\text{cot A}}} \right)}^2}{\text{}} + {\text{}}\left( {{\text{cosec A}} + {\text{cot A}}} \right){\text{}}\left( {{\text{cosec A}} - {\text{cot A}}} \right)}}{{{\text{sec A}}\left( {{\text{cosec A}} - {\text{cot A}}} \right)}}$       $\left[ {\because {\text{}}{{\text{a}}^2} - {{\text{b}}^2} = \left( {{\text{a}} + {\text{b}}} \right)\left( {{\text{a}} - {\text{b}}} \right)} \right]$

$ = \dfrac{{\left( {{\text{cosec A}} - {\text{cot A}}} \right){\text{}}\left[ {\left( {{\text{cosec A}} - {\text{cot A}}} \right){\text{}} + {\text{}}\left( {{\text{cosec A}} - {\text{cot A}}} \right)} \right]}}{{{\text{sec A}}\left( {{\text{cosec A}} - {\text{cot A}}} \right)}}$

$ = \dfrac{{\left[ {\left( {{\text{cosec A}} - {\text{cot A}}} \right){\text{}} + {\text{}}\left( {{\text{cosec A}} - {\text{cot A}}} \right)} \right]}}{{{\text{sec A}}}}$

$ = \dfrac{{2{\text{cosec A}}}}{{{\text{sec A}}}}$

$ = \dfrac{{2{\text{cos A}}}}{{{\text{sin A}}}}$                                                             $\left[ {\because {\text{cosec A}} = \dfrac{1}{{{\text{sin A}}}},{\text{sec A}} = \dfrac{1}{{{\text{cos A}}}}} \right]$

$ = 2{\mathbf{cot}}\;{\mathbf{A}} = {\mathbf{RHS}}$                                                                         $\left[ {\because {\text{cot A}} = \dfrac{{{\text{cos A}}}}{{{\text{sin A}}}}} \right]$                                                     

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

xiii) ${\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}}\left( {\dfrac{{{\mathbf{sec}}\;{\mathbf{A}}\; - \;1}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}} \right) + {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}\left( {\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\; - \;1}}{{1\; + \;{\mathbf{sec}}\;{\mathbf{A}}}}} \right) = 0$

Ans: Taking LHS -

${\text{LHS}} = {\text{co}}{{\text{t}}^2}{\text{A}}\left( {\dfrac{{{\text{sec A}} - {\text{}}1}}{{1{\text{}} + {\text{sin A}}}}} \right) + {\text{se}}{{\text{c}}^2}{\text{A}}\left( {\dfrac{{{\text{sin A}} - {\text{}}1}}{{1{\text{}} + {\text{sec A}}}}} \right)$

$ = {\text{co}}{{\text{t}}^2}{\text{A}}\left( {\dfrac{{{\text{sec A}} - {\text{}}1}}{{1{\text{}} + {\text{sin A}}}} \times \dfrac{{{\text{sec A}} + {\text{}}1}}{{{\text{sec A}} + {\text{}}1}}} \right) + {\text{se}}{{\text{c}}^2}{\text{A}}\left( {\dfrac{{{\text{sin A}} - {\text{}}1}}{{1{\text{}} + {\text{sec A}}}}} \right)$                                                                  

$ = {\text{co}}{{\text{t}}^2}{\text{A}}\left( {\dfrac{{{\text{se}}{{\text{c}}^2}{\text{A}} - {\text{}}1}}{{\left( {1{\text{}} + {\text{sin A}}} \right)\left( {{\text{sec A}} + {\text{}}1} \right)}}} \right) + {\text{se}}{{\text{c}}^2}{\text{A}}\left( {\dfrac{{{\text{sin A}} - {\text{}}1}}{{1{\text{}} + {\text{sec A}}}}} \right)$    $\left[ {\because {\text{}}{{\text{a}}^2} - {{\text{b}}^2} = \left( {{\text{a}} + {\text{b}}} \right)\left( {{\text{a}} - {\text{b}}} \right)} \right]$

$ = {\text{co}}{{\text{t}}^2}{\text{A}}\left( {\dfrac{{{\text{ta}}{{\text{n}}^2}{\text{A}}}}{{\left( {1{\text{}} + {\text{sin A}}} \right)\left( {{\text{sec A}} + {\text{}}1} \right)}}} \right) + {\text{se}}{{\text{c}}^2}{\text{A}}\left( {\dfrac{{{\text{sin A}} - {\text{}}1}}{{1{\text{}} + {\text{sec A}}}}} \right)$              $\left[ {\because {\text{se}}{{\text{c}}^2}{\text{A}} - {\text{ta}}{{\text{n}}^2}{\text{A}} = 1} \right]$

$ = {\text{co}}{{\text{t}}^2}{\text{A}}\left( {\dfrac{{{\text{ta}}{{\text{n}}^2}{\text{A}}}}{{\left( {1{\text{}} + {\text{sin A}}} \right)\left( {{\text{sec A}} + {\text{}}1} \right)}}} \right) + {\text{se}}{{\text{c}}^2}{\text{A}}\left( {\dfrac{{{\text{sin A}} - {\text{}}1}}{{1{\text{}} + {\text{sec A}}}} \times \dfrac{{{\text{sin A}} + {\text{}}1}}{{{\text{sin A}} + {\text{}}1}}} \right)$

$ = \left( {\dfrac{1}{{\left( {1{\text{}} + {\text{sin A}}} \right)\left( {{\text{sec A}} + {\text{}}1} \right)}}} \right) + {\text{se}}{{\text{c}}^2}{\text{A}}\left( {\dfrac{{{\text{sin A}} - {\text{}}1}}{{1{\text{}} + {\text{sec A}}}} \times \dfrac{{{\text{sin A}} + {\text{}}1}}{{{\text{sin A}} + {\text{}}1}}} \right)$                   $\left[ {\because {\text{cot A}} = \dfrac{1}{{{\text{tan A}}}}} \right]$

$ = \left( {\dfrac{1}{{\left( {1{\text{}} + {\text{sin A}}} \right)\left( {{\text{sec A}} + {\text{}}1} \right)}}} \right) + {\text{se}}{{\text{c}}^2}{\text{A}}\left( {\dfrac{{{\text{si}}{{\text{n}}^2}{\text{A}} - {\text{}}1}}{{\left( {1{\text{}} + {\text{sin A}}} \right)\left( {1{\text{}} + {\text{sec A}}} \right)}}} \right)$   

$ = \left( {\dfrac{{1{\text{}} + {\text{se}}{{\text{c}}^2}{\text{A}}\left( { - {\text{co}}{{\text{s}}^2}{\text{A}}} \right)}}{{\left( {1{\text{}} + {\text{sinA}}} \right)\left( {{\text{sec A}} + {\text{}}1} \right)}}} \right)$                                                          $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \left( {\dfrac{{1{\text{}} - {\text{}}1}}{{1{\text{}} + {\text{sin A}}\left( {{\text{sec A}} + {\text{}}1} \right)}}} \right)$                                                                     $\left[ {\because {\text{cos A}} = \dfrac{1}{{{\text{sec A}}}}} \right]$

$ = 0 = {\mathbf{RHS}}$                                                                                                                            

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

xiv) $\dfrac{{{{\left( {1\; - \;2\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}} \right)}^2}}}{{{\mathbf{co}}{{\mathbf{s}}^4}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^4}{\mathbf{A}}}} = 2{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} - 1$

Ans: Taking LHS -

${\text{LHS}} = \dfrac{{{{\left( {1{\text{}} - {\text{}}2{\text{si}}{{\text{n}}^2}{\text{A}}} \right)}^2}}}{{{\text{co}}{{\text{s}}^4}{\text{A}} - {\text{si}}{{\text{n}}^4}{\text{A}}}}$

$ = \dfrac{{{{\left( {{\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} - {\text{}}2{\text{si}}{{\text{n}}^2}{\text{A}}} \right)}^2}}}{{{\text{co}}{{\text{s}}^4}{\text{A}} - {\text{si}}{{\text{n}}^4}{\text{A}}}}$                                                       $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$                                        

$ = \dfrac{{{{\left( {{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; - {\text{si}}{{\text{n}}^2}{\text{A}}} \right)}^2}}}{{{\text{co}}{{\text{s}}^4}{\text{A}} - {\text{si}}{{\text{n}}^4}{\text{A}}}}$     

$ = \dfrac{{{{\left( {{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; - {\text{si}}{{\text{n}}^2}{\text{A}}} \right)}^2}}}{{\left( {{\text{co}}{{\text{s}}^2}{\text{A}} - {\text{si}}{{\text{n}}^2}{\text{A}}} \right)\left( {{\text{co}}{{\text{s}}^2}{\text{A}} + {\text{si}}{{\text{n}}^2}{\text{A}}} \right)}}$                                     $\left[ {\because {\text{}}{{\text{a}}^2} - {{\text{b}}^2} = \left( {{\text{a}} + {\text{b}}} \right)\left( {{\text{a}} - {\text{b}}} \right)} \right]$

$ = {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; - {\text{si}}{{\text{n}}^2}{\text{A}}$                                                              $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; - {\text{}}\left( {1 - {\text{co}}{{\text{s}}^2}{\text{A}}} \right)$                                                  $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = 2{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} - 1 = {\mathbf{RHS}}$                                                                                                                         

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

xv) ${\mathbf{se}}{{\mathbf{c}}^4}{\mathbf{A}}\left( {1 - {\mathbf{si}}{{\mathbf{n}}^4}{\mathbf{A}}} \right) - 2{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}} = 1$

Ans: Taking LHS -

${\text{LHS}} = {\text{se}}{{\text{c}}^4}{\text{A}}\left( {1 - {\text{si}}{{\text{n}}^4}{\text{A}}} \right) - 2{\text{ta}}{{\text{n}}^2}{\text{A}}$

$ = {\text{se}}{{\text{c}}^4}{\text{A}}\left( {1 - {\text{si}}{{\text{n}}^2}{\text{A}}} \right)\left( {1 + {\text{si}}{{\text{n}}^2}{\text{A}}} \right) - 2{\text{ta}}{{\text{n}}^2}{\text{A}}$      $\left[ {\because {\text{}}{{\text{a}}^2} - {{\text{b}}^2} = \left( {{\text{a}} + {\text{b}}} \right)\left( {{\text{a}} - {\text{b}}} \right)} \right]$                                        

$ = {\text{se}}{{\text{c}}^4}{\text{A}}\left( {{\text{co}}{{\text{s}}^2}{\text{A}}} \right)\left( {1 + {\text{si}}{{\text{n}}^2}{\text{A}}} \right) - 2{\text{ta}}{{\text{n}}^2}{\text{A}}$                           $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = {\text{se}}{{\text{c}}^2}{\text{A}}\left( {1 + {\text{si}}{{\text{n}}^2}{\text{A}}} \right) - 2{\text{ta}}{{\text{n}}^2}{\text{A}}$                                        $\left[ {\because {\text{}}{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}} \times {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = {\text{se}}{{\text{c}}^2}{\text{A}} + {\text{si}}{{\text{n}}^2}{\text{A}}.{\text{se}}{{\text{c}}^2}{\text{A}} - 2{\text{ta}}{{\text{n}}^2}{\text{A}}$                                    

$ = {\text{se}}{{\text{c}}^2}{\text{A}} + \dfrac{{{\text{si}}{{\text{n}}^2}{\text{A}}}}{{{\text{co}}{{\text{s}}^2}{\text{A}}}} - 2{\text{ta}}{{\text{n}}^2}{\text{A}}$ 

$ = {\text{se}}{{\text{c}}^2}{\text{A}} + {\text{ta}}{{\text{n}}^2}{\text{A}} - 2{\text{ta}}{{\text{n}}^2}{\text{A}}$     

$ = {\text{se}}{{\text{c}}^2}{\text{A}} - {\text{ta}}{{\text{n}}^2}{\text{A}}$                                                 

$ = 1 = {\mathbf{RHS}}$                                                                       $\left[ {\because {\text{se}}{{\text{c}}^2}{\text{A}} - {\text{ta}}{{\text{n}}^2}{\text{A}} = 1} \right]$                                                            

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

xvi) ${\mathbf{cose}}{{\mathbf{c}}^4}{\mathbf{A}}\left( {1 - {\mathbf{co}}{{\mathbf{s}}^4}{\mathbf{A}}} \right) - 2{\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} = 1$

Ans: Taking LHS -

${\text{LHS}} = {\text{cose}}{{\text{c}}^4}{\text{A}}\left( {1 - {\text{co}}{{\text{s}}^4}{\text{A}}} \right) - 2{\text{co}}{{\text{t}}^2}{\text{A}}$

$ = {\text{cose}}{{\text{c}}^4}{\text{A}}\left( {1 - {\text{co}}{{\text{s}}^2}{\text{A}}} \right)\left( {1 + {\text{co}}{{\text{s}}^2}{\text{A}}} \right) - 2{\text{co}}{{\text{t}}^2}{\text{A}}$  $\left[ {\because {\text{}}{{\text{a}}^2} - {{\text{b}}^2} = \left( {{\text{a}} + {\text{b}}} \right)\left( {{\text{a}} - {\text{b}}} \right)} \right]$                                        

$ = {\text{cose}}{{\text{c}}^4}{\text{A}}\left( {{\text{si}}{{\text{n}}^2}{\text{A}}} \right)\left( {1 + {\text{co}}{{\text{s}}^2}{\text{A}}} \right) - 2{\text{co}}{{\text{t}}^2}{\text{A}}$                       $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = {\text{cose}}{{\text{c}}^2}{\text{A}}\left( {1 + {\text{co}}{{\text{s}}^2}{\text{A}}} \right) - 2{\text{co}}{{\text{t}}^2}{\text{A}}$                                 $\left[ {\because {\text{co}}{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}} \times {\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} = 1} \right]$

$ = {\text{cose}}{{\text{c}}^2}{\text{A}} + {\text{co}}{{\text{s}}^2}{\text{A}}.{\text{cose}}{{\text{c}}^2}{\text{A}} - 2{\text{co}}{{\text{t}}^2}{\text{A}}$                                    

$ = {\text{cose}}{{\text{c}}^2}{\text{A}} + \dfrac{{{\text{co}}{{\text{s}}^2}{\text{A}}}}{{{\text{si}}{{\text{n}}^2}{\text{A}}}} - 2{\text{co}}{{\text{t}}^2}{\text{A}}$    

$ = {\text{cose}}{{\text{c}}^2}{\text{A}} + {\text{co}}{{\text{t}}^2}{\text{A}} - 2{\text{co}}{{\text{t}}^2}{\text{A}}$                               

$ = {\text{cose}}{{\text{c}}^2}{\text{A}} - {\text{co}}{{\text{t}}^2}{\text{A}}$                              

$ = 1 = {\mathbf{RHS}}$                                                                   $\left[ {\because {\text{cose}}{{\text{c}}^2}{\text{A}} - {\text{co}}{{\text{t}}^2}{\text{A}} = 1} \right]$                                                                         

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

xvii) $\left( {1 + {\mathbf{tan}}\;{\mathbf{A}} + {\mathbf{sec}}\;{\mathbf{A}}} \right)\left( {1 + {\mathbf{cot}}\;{\mathbf{A}} - {\mathbf{cosec}}\;{\mathbf{A}}} \right) = 2$

Ans: Taking LHS -

${\text{LHS}} = \left( {1 + {\text{tan A}} + {\text{sec A}}} \right)\left( {1 + {\text{cot A}} - {\text{cosec A}}} \right)$

$ = \left( {1 + \dfrac{{{\text{sin A}}}}{{{\text{cos A}}}} + \dfrac{1}{{{\text{cos A}}}}} \right)\left( {1 + \dfrac{{{\text{cos A}}}}{{{\text{sin A}}}} - \dfrac{1}{{{\text{sin A}}}}} \right)$         

$\left[ {\because {\text{ cos A}} = \dfrac{1}{{{\text{sec A}}}},{\text{ sin A}} = \dfrac{1}{{{\text{cosec A}}}},{\text{ tan A}} = \dfrac{{{\text{sinA}}}}{{{\text{cos A}}}},{\text{cot A}} = \dfrac{{{\text{cosA}}}}{{{\text{sin A}}}}} \right]$

$ = \left( {\dfrac{{{\text{cos A}} + {\text{ sin A}} + {\text{}}1}}{{{\text{cos A}}}}} \right)\left( {\dfrac{{{\text{sin A}} + {\text{ cos A}} - 1}}{{{\text{sin A}}}}} \right)$                   

$ = \left( {\dfrac{{{{({\text{cos A }} + {\text{ sin A}})}^2}{\text{}} - {\text{}}{1^2}}}{{{\text{cos A}}.{\text{sin A}}}}} \right)$                                              $\left[ {\because {\text{}}{{\text{a}}^2} - {{\text{b}}^2} = \left( {{\text{a}} + {\text{b}}} \right)\left( {{\text{a}} - {\text{b}}} \right)} \right]$

$ = \left( {\dfrac{{{\text{co}}{{\text{s}}^2}{\text{A}} + {\text{si}}{{\text{n}}^2}{\text{A}} + {\text{}}2{\text{sin A}}.{\text{cos A}} - {\text{}}1}}{{{\text{cos A}}.{\text{sin A}}}}} \right)$                               $\left[ {\because {\text{}}{{\text{a}}^2} + {{\text{b}}^2} + 2{\text{ab}} = {{({\text{a}} + {\text{b}})}^2}} \right]$

$ = \left( {\dfrac{{1{\text{}} + {\text{}}2{\text{sin A}}.{\text{cos A}} - {\text{}}1}}{{{\text{cos A}}.{\text{sin A}}}}} \right)$                                                            $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \left( {\dfrac{{2{\text{sin A}}.{\text{cos A}}}}{{{\text{cos A}}.{\text{sin A}}}}} \right)$                               

$ = 1 = {\mathbf{RHS}}$                                                                                                                                   

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

2. If ${\mathbf{sin}}\;{\mathbf{A}} + {\mathbf{cos}}\;{\mathbf{A}} = {\mathbf{p}}$ and ${\mathbf{sec}}\;{\mathbf{A}} + {\mathbf{cosec}}\;{\mathbf{A}} = {\mathbf{q}},$ then prove that: ${\mathbf{q}}\left( {{{\mathbf{p}}^2} - 1} \right) = 2{\mathbf{p}}$.

Ans: Taking LHS -

${\text{LHS}} = {\text{q}}\left( {{{\text{p}}^2} - 1} \right)$

$ = \left( {{\text{sec A}} + {\text{cosec A}}} \right)\left[ {{{\left( {{\text{sin A}} + {\text{cos A}}} \right)}^2} - 1} \right]$         

$ = \left( {{\text{sec A}} + {\text{cosec A}}} \right)\left[ {{\text{si}}{{\text{n}}^2}{\text{A}} + {\text{co}}{{\text{s}}^2}{\text{A}} + 2{\text{sin A}}.{\text{cosA}} - 1} \right]$     

                                                               $\left[ {\because {\text{}}{{\text{a}}^2} + {{\text{b}}^2} + 2{\text{ab}} = {{({\text{a}} + {\text{b}})}^2}} \right]$              

$ = \left( {{\text{sec A}} + {\text{cosec A}}} \right)\left[ {1 + 2{\text{sin A}}.{\text{cosA}} - 1} \right]$                 $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \left( {{\text{sec A}} + {\text{cosec A}}} \right)\left[ {2{\text{sin A}}.{\text{cosA}}} \right]$

$ = \left[ {2{\text{sin A}}.{\text{cosA}}.{\text{sec A}} + 2{\text{sin A}}.{\text{cosA}}.{\text{cosec A}}} \right]$

$\left[{\because{\text{cosA}}=\dfrac{1}{{{\text{secA}}}},{\text{sinA}}=\dfrac{1}{{{\text{cosec A}}}}} \right]$

$ = 2{\text{sin A}} + 2{\text{cos A}}$

$ = 2{\mathbf{p}} = {\mathbf{RHS}}$                                                                                                                                   

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

3. If ${\mathbf{x}} = {\mathbf{acos}}\;{\mathbf{\theta }}$ and ${\mathbf{y}} = {\mathbf{bcot}}\;{\mathbf{\theta }}$, show that:

$\dfrac{{{{\mathbf{a}}^2}}}{{{{\mathbf{x}}^2}}} - \dfrac{{{{\mathbf{b}}^2}}}{{{{\mathbf{y}}^2}}} = 1$

Ans: Taking LHS -

${\text{LHS}} = \dfrac{{{{\text{a}}^2}}}{{{{\text{x}}^2}}} - \dfrac{{{{\text{b}}^2}}}{{{{\text{y}}^2}}}$

${\mathbf{x}} = {\mathbf{acos}}\;{\mathbf{\theta }}\;\& \;{\mathbf{y}} = {\mathbf{bcot}}\;{\mathbf{\theta }}$ 

$ \Rightarrow \dfrac{{\mathbf{x}}}{{\mathbf{a}}} = {\mathbf{cos}}\;{\mathbf{\theta }}\; \& \;\dfrac{{\mathbf{y}}}{{\mathbf{b}}} = {\mathbf{cot}}\;{\mathbf{\theta }}$

$ \Rightarrow \dfrac{{\mathbf{a}}}{{\mathbf{x}}} = {\mathbf{sec}}\;{\mathbf{\theta }}\;\& \;\dfrac{{\mathbf{b}}}{{\mathbf{y}}} = {\mathbf{tan}}\;{\mathbf{\theta }}$                            $\left[ {\because {\text{cosA}} = \dfrac{1}{{{\text{sec A}}}},{\text{ tan A}} = \dfrac{1}{{{\text{cot A}}}}} \right]$

So, 

$\dfrac{{{{\text{a}}^2}}}{{{{\text{x}}^2}}} - \dfrac{{{{\text{b}}^2}}}{{{{\text{y}}^2}}} = {\text{se}}{{\text{c}}^2}{{\theta}} - {\text{ta}}{{\text{n}}^2}{{\theta}}$

$\dfrac{{{{\text{a}}^2}}}{{{{\text{x}}^2}}} - \dfrac{{{{\text{b}}^2}}}{{{{\text{y}}^2}}} = 1$                                                                    $\left[ {\because {\text{se}}{{\text{c}}^2}{\text{A}} - {\text{ta}}{{\text{n}}^2}{\text{A}} = 1} \right]$ 

$ = 1 = {\mathbf{RHS}}$                                                                                                                                   

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

4. If ${\mathbf{sec}}\;{\mathbf{A}} + {\mathbf{tan}}\;{\mathbf{A}} = {\mathbf{p}},\;$show that:

${\mathbf{sin}}\;{\mathbf{A}} = \dfrac{{{{\mathbf{p}}^2}\; - \;1}}{{{{\mathbf{p}}^2}\; + \;1}}$

Ans: Taking RHS -

${\text{RHS}} = \dfrac{{{{\mathbf{p}}^2}\; - \;1}}{{{{\mathbf{p}}^2}\; + \;1}}$

${\mathbf{sec}}\;{\mathbf{A}} + {\mathbf{tan}}\;{\mathbf{A}} = {\mathbf{p}}$ 

$ = \dfrac{{{{\left( {{\mathbf{sec}}\;{\mathbf{A}} + {\mathbf{tan}}\;{\mathbf{A}}} \right)}^2}\; - \;1}}{{{{\left( {{\mathbf{sec}}\;{\mathbf{A}} + {\mathbf{tan}}\;{\mathbf{A}}} \right)}^2}\; + \;1}}$

$ = \dfrac{{{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}\; + \;{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;2\;{\mathbf{sec}}\;{\mathbf{A}}\;.\;{\mathbf{tan}}\;{\mathbf{A}}\; - \;1}}{{{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}\; + \;{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;2\;{\mathbf{sec}}\;{\mathbf{A}}\;.\;{\mathbf{tan}}\;{\mathbf{A}}\; + \;1}}$                     

$ = \dfrac{{{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;2\;{\mathbf{sec}}\;{\mathbf{A}}\;.\;{\mathbf{tan}}\;{\mathbf{A}}}}{{{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}\; + \;{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}\; + \;2\;{\mathbf{sec}}\;{\mathbf{A}}\;.\;{\mathbf{tan}}\;{\mathbf{A}}}}$                                                $\left[ {\because {\text{se}}{{\text{c}}^2}{\text{A}} = {\text{ta}}{{\text{n}}^2}{\text{A}} + 1} \right]$  

$ = \dfrac{{2\;{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;2\;{\mathbf{sec}}\;{\mathbf{A}}\;.\;{\mathbf{tan}}\;{\mathbf{A}}}}{{2\;{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}\; + \;2\;{\mathbf{sec}}\;{\mathbf{A}}\;.\;{\mathbf{tan}}\;{\mathbf{A}}}}$ 

$ = \dfrac{{2\;{\mathbf{tan}}\;{\mathbf{A}}\;\left( {{\mathbf{tan}}\;{\mathbf{A}}\; + \;{\mathbf{sec}}\;{\mathbf{A}}} \right)}}{{2\;{\mathbf{sec}}\;{\mathbf{A}}\;\left( {{\mathbf{sec}}\;{\mathbf{A}}\; + \;{\mathbf{tan}}\;{\mathbf{A}}} \right)}}$

$ = \dfrac{{{\mathbf{tan}}\;{\mathbf{A}}}}{{{\mathbf{sec}}\;{\mathbf{A}}}}$

$ = \dfrac{{\dfrac{{{\text{sin A}}}}{{{\text{cos A}}}}}}{{\dfrac{1}{{{\text{cos A}}}}}}$                                                                $\left[ {\because {\text{sec A}} = \dfrac{1}{{{\text{cos A}}}},{\text{tan A}} = \dfrac{{{\text{sin A}}}}{{{\text{cos A}}}}} \right]$

$ = {\text{sin A}} = {\text{LHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

5. If ${\mathbf{tan}}\;{\mathbf{A}} = {\mathbf{n}}.{\mathbf{tan}}\;{\mathbf{B}}$ and ${\mathbf{sin}}\;{\mathbf{A}} = {\mathbf{m}}.\;{\mathbf{sin}}\;{\mathbf{B}}$, prove that: ${\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = \dfrac{{{{\mathbf{m}}^2}\; - \;1}}{{{{\mathbf{n}}^2}\; - \;1}}$.

Ans: Taking RHS -

${\text{RHS}} = \dfrac{{{{\mathbf{m}}^2}\; - \;1}}{{{{\mathbf{n}}^2}\; - \;1}}$

${\mathbf{tan}}\;{\mathbf{A}} = {\mathbf{n}}.{\mathbf{tan}}\;{\mathbf{B}}\;\;\& \;\;{\mathbf{sin}}\;{\mathbf{A}} = {\mathbf{m}}.\;{\mathbf{sin}}\;{\mathbf{B}}$ 

$ \Rightarrow {\mathbf{n}} = \dfrac{{{\mathbf{tan}}\;{\mathbf{A}}}}{{{\mathbf{tan}}\;{\mathbf{B}}}}\;\;\& \;\;{\mathbf{m}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{B}}}}$

$ = \dfrac{{{{\left( {\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{B}}}}} \right)}^2}\; - \;1}}{{{{\left( {\dfrac{{{\mathbf{tan}}\;{\mathbf{A}}}}{{{\mathbf{tan}}\;{\mathbf{B}}}}} \right)}^2}\; - \;1}}$

$ = \dfrac{{{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{B}}.\left( {{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}} \right)}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}.\left( {{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{B}}} \right)}}$                     

$ = \dfrac{{\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}}}.\left( {{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}} \right)}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}.\left( {\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}\; - \;\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}}}} \right)}}$                                                                        $\left[ {\because {\text{tan A}} = \dfrac{{{\text{sin A}}}}{{{\text{cos A}}}}} \right]$  

$ = \dfrac{{\left( {{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}} \right)}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}.\left( {\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}\; - \;\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}}}} \right)}}$ 

Multiply numerator and denominator by ${\text{co}}{{\text{s}}^2}{\text{A}}.$

$ = \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}.\left( {{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}} \right)}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}.\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}.\left( {\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}\; - \;\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}}}} \right)}}$

$ = \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}.\left( {{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}} \right)}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}\;.\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\;.\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}}$

$ = \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}.\left( {1\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; - \;1\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}} \right)}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}\;.\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\;.\;\left( {1\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}} \right)}}$                                              $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}.\left( {{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\;} \right)}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}\;\left( {{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}} \right)\;\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}.\left( {{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\;} \right)}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}\;\left( 1 \right)\;\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$                                                            $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = {\mathbf{LHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

6. If $2\;{\mathbf{sin}}\;{\mathbf{A}} - 1 = 0,$ show that: ${\mathbf{sin}}\;3{\mathbf{A}} = 3{\mathbf{sin}}\;{\mathbf{A}} - 4{\mathbf{si}}{{\mathbf{n}}^3}{\mathbf{A}}$.

Ans: Given $2{\text{sin A}} - 1 = 0$

$ \Rightarrow {\text{sin A}} = \dfrac{1}{2}$

We know that ${\text{sin}}{30^ \circ } = \dfrac{1}{2}$

$ \Rightarrow {\text{sin A}} = {\text{sin }}{30^ \circ }$

$ \Rightarrow {\text{A}} = {30^ \circ }$

Taking LHS -

${\text{LHS}} = {\text{sin}}3{\text{A}}$

$ = {\text{sin }}{90^ \circ }$

$ = 1$

Taking RHS - 

${\text{RHS}} = 3{\text{sin A}} - 4{\text{si}}{{\text{n}}^3}{\text{A}}$

$ = 3\left( {\dfrac{1}{2}} \right) - 4{\left( {\dfrac{1}{2}} \right)^3}$

$ = \dfrac{3}{2} - \dfrac{4}{8}$

$ = 1 = {\text{LHS}}$

$ \Rightarrow {\text{LHS}} = {\text{RHS}}$

Hence proved.

7. Evaluate: 

i) $2{\left( {\dfrac{{{\text{tan}}{{35}^ \circ }}}{{{\text{cot}}{{55}^ \circ }}}} \right)^2} + {\left( {\dfrac{{{\text{cot}}{{55}^ \circ }}}{{{\text{tan}}{{35}^ \circ }}}} \right)^2} - 3{\left( {\dfrac{{{\text{sec}}{{40}^ \circ }}}{{{\text{cosec}}{{50}^ \circ }}}} \right)^{}}$

Ans: $2{\left( {\dfrac{{{\text{tan}}{{35}^ \circ }}}{{{\text{cot}}{{55}^ \circ }}}} \right)^2} + {\left( {\dfrac{{{\text{cot}}{{55}^ \circ }}}{{{\text{tan}}{{35}^ \circ }}}} \right)^2} - 3{\left( {\dfrac{{{\text{sec}}{{40}^ \circ }}}{{{\text{cosec}}{{50}^ \circ }}}} \right)^{}}$

$ = 2{\left( {\dfrac{{{\text{tan}}{{\left( {90 - 55} \right)}^ \circ }}}{{{\text{cot}}{{55}^ \circ }}}} \right)^2} + {\left( {\dfrac{{{\text{cot}}{{55}^ \circ }}}{{{\text{tan}}{{\left( {90 - 55} \right)}^ \circ }}}} \right)^2} - 3{\left( {\dfrac{{{\text{sec}}{{\left( {90 - 50} \right)}^ \circ }}}{{{\text{cosec}}{{50}^ \circ }}}} \right)^{}}$

$ = 2{\left( {\dfrac{{{\text{cot}}{{55}^ \circ }}}{{{\text{cot}}{{55}^ \circ }}}} \right)^2} + {\left( {\dfrac{{{\text{cot}}{{55}^ \circ }}}{{{\text{cot}}{{55}^ \circ }}}} \right)^2} - 3{\left( {\dfrac{{{\mathbf{co}}{\text{sec}}{{50}^ \circ }}}{{{\text{cosec}}{{50}^ \circ }}}} \right)^{}}$        $\left[ {\because {\text{tan}}\left( {90 - {\text{A}}} \right) = {\text{cot A}}{\& ^{}}{\text{sec}}\left( {90 - {\text{A}}} \right) = {\text{cosec}}{{\text{A}}^{}}} \right]$

$ = 2\left( 1 \right) + 1 - 3\left( 1 \right)$

$ = 3 - 3 = 0$

ii) ${\mathbf{sec}}{\text{}}{26^ \circ }{\text{sin}}{64^ \circ } + \dfrac{{{\mathbf{co}}{\text{sec}}{{33}^ \circ }}}{{{\text{sec}}{{57}^ \circ }}}$

Ans: ${\mathbf{sec}}{\text{}}{26^ \circ }{\text{sin}}{64^ \circ } + \dfrac{{{\mathbf{co}}{\text{sec}}{{33}^ \circ }}}{{{\text{sec}}{{57}^ \circ }}}$

$ = {\mathbf{sec}}{\text{}}{\left( {90 - 64} \right)^ \circ }{\text{sin}}{64^ \circ } + \dfrac{{{\mathbf{co}}{\text{sec}}{{33}^ \circ }}}{{{\text{sec}}{{\left( {90 - 33} \right)}^ \circ }}}$

$ = {\mathbf{cosec}}{\text{}}{64^ \circ }{\text{sin}}{64^ \circ } + \dfrac{{{\mathbf{co}}{\text{sec}}{{33}^ \circ }}}{{{\mathbf{co}}{\text{sec}}{{33}^ \circ }}}$                             $\left[ {\because {\text{sec}}\left( {90 - {\text{A}}} \right) = {\text{cosec}}{{\text{A}}^{}}} \right]$

$ = 1 + 1$                                                                         $\left[ {\because {\text{sin A}}.{\text{cosec A}} = {1^{}}} \right] = 2$

iii) $\dfrac{{5\;{\mathbf{sin}}\;{\text{}}{{66}^ \circ }}}{{{\mathbf{cos}}\;{{24}^ \circ }}} - \dfrac{{2\;{\mathbf{cot}}\;{\text{}}{{85}^ \circ }}}{{{\mathbf{tan}}\;{5^ \circ }}}$

Ans: $\dfrac{{5\;{\mathbf{sin}}\;{\text{}}{{66}^ \circ }}}{{{\mathbf{cos}}\;{{24}^ \circ }}} - \dfrac{{2\;{\mathbf{cot}}\;{\text{}}{{85}^ \circ }}}{{{\mathbf{tan}}\;{5^ \circ }}}$

$ = \dfrac{{5\;{\mathbf{sin}}\;{\text{}}{{66}^ \circ }}}{{{\mathbf{cos}}\;{{\left( {90 - 66} \right)}^ \circ }}} - \dfrac{{2\;{\mathbf{cot}}\;{\text{}}{{85}^ \circ }}}{{{\mathbf{tan}}\;{{\left( {90 - 85} \right)}^ \circ }}}$

$ = \dfrac{{5\;{\mathbf{sin}}\;{\text{}}{{66}^ \circ }}}{{{\mathbf{sin}}\;{{66}^ \circ }}} - \dfrac{{2\;{\mathbf{cot}}\;{\text{}}{{85}^ \circ }}}{{{\mathbf{cot}}\;{{85}^ \circ }}}$      $\left[ {\because {\text{sin}}\left( {90 - {\text{A}}} \right) = {\text{cos A}} \;\&\; {\text{tan}}\left( {90 - {\text{A}}} \right) = {\text{cot A}}} \right]$

$ = 5 - 2$

$ = 3$

iv) ${\mathbf{cos}}\;{\text{}}{40^ \circ }{\text{cosec}}{50^ \circ } + {\text{sin}}{50^ \circ }{\text{sec}}{40^ \circ }$

Ans: ${\mathbf{cos}}\;{\text{}}{40^ \circ }{\text{cosec}}{50^ \circ } + {\text{sin}}{50^ \circ }{\text{sec}}{40^ \circ }$

$ = {\mathbf{cos}}\;{\text{}}{40^ \circ }{\text{cosec}}{\left( {90 - 40} \right)^ \circ } + {\text{sin}}{50^ \circ }{\text{sec}}{\left( {90 - 50} \right)^ \circ }$

$ = {\mathbf{cos}}\;{\text{}}{40^ \circ }{\text{sec}}{40^ \circ } + {\text{sin}}{50^ \circ }{\text{cosec}}{50^ \circ }$     $\left[ {\because {\text{sec}}\left( {90 - {\text{A}}} \right) = {\text{cosec}}{{\text{A}}^{}}\;\; \&\;\; {\text{cosec}}\left( {90 - {\text{A}}} \right) = {\text{sec}}{{\text{A}}^{}}} \right]$

$ = 1 + 1$                      $\left[ {\because {\text{secA}}.{\text{cos A}} = {1^{}}\;\; \&\;\; {\text{cosec A}}.{\text{sin A}} = 1} \right]$

$ = 2$

v) ${\mathbf{sin}}\;{\text{}}{27^ \circ }{\text{sin}}{63^ \circ } - {\text{cos}}{63^ \circ }{\text{cos}}{27^ \circ }$

Ans: ${\mathbf{sin}}\;{\text{}}{27^ \circ }{\text{sin}}{63^ \circ } - {\text{cos}}{63^ \circ }{\text{cos}}{27^ \circ }$

$ = {\mathbf{sin}}\;{\text{}}{27^ \circ }{\text{sin}}{\left( {90 - 27} \right)^ \circ } - {\text{cos}}{\left( {90 - 27} \right)^ \circ }{\text{cos}}{27^ \circ }$

$ = {\mathbf{sin}}\;{\text{}}{27^ \circ }.{\text{cos}}{27^ \circ } - {\mathbf{sin}}\;{\text{}}{27^ \circ }.{\text{cos}}{27^ \circ }$       $\left[ {\because {\text{sin}}\left( {90 - {\text{A}}} \right) = {\text{cos A}}\;\&\; {\text{cos}}\left( {90 - {\text{A}}} \right) = {\text{sin A}}} \right]$

$ = 0$

vi) $\dfrac{{3\;{\mathbf{sin}}\;{\text{}}{{72}^ \circ }}}{{{\mathbf{cos}}\;{{18}^ \circ }}} - \dfrac{{\;{\mathbf{sec}}\;{\text{}}{{32}^ \circ }}}{{{\mathbf{cosec}}\;{{58}^ \circ }}}$

Ans: $\dfrac{{3\;{\mathbf{sin}}\;{\text{}}{{72}^ \circ }}}{{{\mathbf{cos}}\;{{18}^ \circ }}} - \dfrac{{\;{\mathbf{sec}}\;{\text{}}{{32}^ \circ }}}{{{\mathbf{cosec}}\;{{58}^ \circ }}}$

$ = \dfrac{{3\;{\mathbf{sin}}\;{\text{}}{{72}^ \circ }}}{{{\mathbf{cos}}\;{{\left( {90 - 72} \right)}^ \circ }}} - \dfrac{{\;{\mathbf{sec}}\;{\text{}}{{32}^ \circ }}}{{{\mathbf{cosec}}\;{{\left( {90 - 32} \right)}^ \circ }}}$

$ = \dfrac{{3\;{\mathbf{sin}}\;{\text{}}{{72}^ \circ }}}{{{\mathbf{sin}}\;{\text{}}{{72}^ \circ }}} - \dfrac{{\;{\mathbf{sec}}\;{\text{}}{{32}^ \circ }}}{{{\mathbf{sec}}\;{{32}^ \circ }}}$     $\left[ {\because {\text{cosec}}\left( {90 - {\text{A}}} \right) = {\text{sec}}{{\text{A}}^{}} \;\&\; {\text{cos}}\left( {90 - {\text{A}}} \right) = {\text{sin}}{{\text{A}}^{}}} \right]$

$ = 3\left( 1 \right) - 1$

$ = 2$

vii) $3{\mathbf{cos}}\;{\text{}}{80^ \circ }{\text{cosec}}{10^ \circ } + 2{\text{cos}}{59^ \circ }{\text{cosec}}{31^ \circ }$

Ans: $3{\mathbf{cos}}\;{\text{}}{80^ \circ }{\text{cosec}}{10^ \circ } + 2{\text{cos}}{59^ \circ }{\text{cosec}}{31^ \circ }$

$ = 3{\mathbf{cos}}{\text{}}{80^ \circ }{\text{cosec}}{\left( {90 - 80} \right)^ \circ } + 2{\mathbf{cos}}{\text{}}{59^ \circ }{\text{cosec}}{\left( {90 - 59} \right)^ \circ }$

$ = 3{\mathbf{cos}}{\text{}}{80^ \circ }{\text{sec}}{80^ \circ } + 2{\mathbf{cos}}{\text{}}{59^ \circ }{\text{sec}}{59^ \circ }$             $\left[ {\because {\text{cosec}}\left( {90 - {\text{A}}} \right) = {\text{sec}}{{\text{A}}^{}}} \right]$

$ = 3\left( 1 \right) + 2\left( 1 \right)$                                                                      $\left[ {\because {\text{sec A}}.{\text{cos A}} = {1^{}}} \right]$

$ = 5$

viii) $\dfrac{{{\mathbf{cos}}\;{\text{}}{{75}^ \circ }}}{{{\mathbf{sin}}\;{{15}^ \circ }}} + \dfrac{{{\mathbf{sin}}\;{\text{}}{{12}^ \circ }}}{{{\mathbf{cos}}\;{{78}^ \circ }}} - \dfrac{{{\mathbf{cos}}\;{\text{}}{{18}^ \circ }}}{{{\mathbf{sin}}\;{{72}^ \circ }}}$

Ans: $\dfrac{{{\mathbf{cos}}\;{\text{}}{{75}^ \circ }}}{{{\mathbf{sin}}\;{{15}^ \circ }}} + \dfrac{{{\mathbf{sin}}\;{\text{}}{{12}^ \circ }}}{{{\mathbf{cos}}\;{{78}^ \circ }}} - \dfrac{{{\mathbf{cos}}\;{\text{}}{{18}^ \circ }}}{{{\mathbf{sin}}\;{{72}^ \circ }}}$

$ = \dfrac{{{\mathbf{cos}}\;{\text{}}{{\left( {90 - 15} \right)}^ \circ }}}{{{\mathbf{sin}}\;{{15}^ \circ }}} + \dfrac{{{\mathbf{sin}}\;{\text{}}{{12}^ \circ }}}{{{\mathbf{cos}}\;{{\left( {90 - 12} \right)}^ \circ }}} - \dfrac{{{\mathbf{cos}}\;{\text{}}{{\left( {90 - 72} \right)}^ \circ }}}{{{\mathbf{sin}}\;{{72}^ \circ }}}$

$ = \dfrac{{{\mathbf{sin}}\;{{15}^ \circ }}}{{{\mathbf{sin}}\;{{15}^ \circ }}} + \dfrac{{{\mathbf{sin}}\;{\text{}}{{12}^ \circ }}}{{{\mathbf{sin}}\;{{12}^ \circ }}} - \dfrac{{{\mathbf{sin}}\;{{72}^ \circ }}}{{{\mathbf{sin}}\;{{72}^ \circ }}}$                                       $\left[ {\because {\text{cos}}\left( {90 - {\text{A}}} \right) = {\text{sin}}{{\text{A}}^{}}} \right]$

$ = 1 + 1 - 1$

$ = 1$

8. Prove that:

i) ${\mathbf{tan}}\left( {{{55}^ \circ } + {\text{x}}} \right) = {\text{cot}}\left( {{{35}^ \circ } - {\text{x}}} \right)$

Ans: Taking LHS -

${\text{LHS}} = {\text{tan}}\left( {{{55}^ \circ } + {\text{x}}} \right)$

$ = {\text{tan}}\left[ {{{\left( {90 - 35} \right)}^ \circ } + {\text{x}})} \right]$

$ = {\text{tan}}\left[ {{{90}^ \circ } - \left( {{{35}^ \circ } - {\text{x}}} \right)} \right]$

$ = {\text{cot}}\left( {{{35}^ \circ } - {\text{x}}} \right) = {\text{RHS}}$                                             $\left[ {\because {\text{tan}}\left( {90 - {\text{A}}} \right) = {\text{cot}}{{\text{A}}^{}}} \right]$

$ \Rightarrow {\mathbf{LHS}} = {\mathbf{RHS}}$

Hence proved.

ii) ${\mathbf{sec}}\left( {{{70}^ \circ } - {{\theta }}} \right) = {\text{cosec}}\left( {{{20}^ \circ } + {{\theta }}} \right)$

Ans: Taking LHS -

${\text{LHS}} = {\text{sec}}\left( {{{70}^ \circ } - {{\theta }}} \right)$

$ = {\text{sec}}\left( {{{\left( {90 - 20} \right)}^ \circ } - {{\theta }}} \right)$

$ = {\text{sec}}\left[ {{{90}^ \circ } - \left( {{{20}^ \circ } + {{\theta }}} \right)} \right]$

$ = {\text{cosec}}\left( {{{20}^ \circ } + {{\theta }}} \right) = {\text{RHS}}$                                     $\left[ {\because {\text{sec}}\left( {90 - {\text{A}}} \right) = {\text{cosec}}{{\text{A}}^{}}} \right]$

$ \Rightarrow {\mathbf{LHS}} = {\mathbf{RHS}}$

Hence proved.

iii) ${\mathbf{sin}}\left( {{{28}^ \circ } + {\text{A}}} \right) = {\text{cos}}\left( {{{62}^ \circ } - {\text{A}}} \right)$

Ans: Taking LHS -

${\text{LHS}} = {\text{sin}}\left( {{{28}^ \circ } + {\text{A}}} \right)$

$ = {\text{sin}}\left[ {{{\left( {90 - 62} \right)}^ \circ } + {\text{A}}} \right]$

$ = {\text{sin}}\left[ {{{90}^ \circ } - \left( {{{62}^ \circ } - {\text{A}}} \right)} \right]$

$ = {\text{cos}}\left( {{{62}^ \circ } - {\text{A}}} \right) = {\text{RHS}}$                                            $\left[ {\because {\text{sin}}\left( {90 - {\text{A}}} \right) = {\text{cos}}{{\text{A}}^{}}} \right]$

$ \Rightarrow {\mathbf{LHS}} = {\mathbf{RHS}}$

Hence proved.

iv) $\dfrac{1}{{1\; + \;{\mathbf{cos}}\;\left( {{{90}^ \circ } - {\text{A}}} \right)}} + \dfrac{1}{{1\; - \;{\mathbf{cos}}\;\left( {{{90}^ \circ } - {\text{A}}} \right)}} = 2{\mathbf{cose}}{{\mathbf{c}}^2}\left( {{{90}^ \circ } - {\text{A}}} \right)$

Ans:  Taking LHS -

$ = \dfrac{1}{{1\; + \;{\mathbf{cos}}\;\left( {{{90}^ \circ } - {\text{A}}} \right)}} + \dfrac{1}{{1\; - \;{\mathbf{cos}}\;\left( {{{90}^ \circ } - {\text{A}}} \right)}}$

$ = \dfrac{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}} + \left( {1\; + {\mathbf{sin}}{\text{A}}} \right)}}{{\left( {1{\text{}} + {\text{sin A}}} \right)\left( {1\; - \;{\text{sin A}}} \right)}}$                                                       $\left[ {\because {\text{cos}}\left( {90 - {\text{A}}} \right) = {\text{sin}}{{\text{A}}^{}}} \right]$

$ = \dfrac{2}{{{1^2} - {\text{si}}{{\text{n}}^2}{\text{A}}}}$                                                               $\left[ {\because {\text{}}{{\text{a}}^2} - {{\text{b}}^2} = \left( {{\text{a}} + {\text{b}}} \right)\left( {{\text{a}} - {\text{b}}} \right)} \right]$

$ = \dfrac{2}{{{\text{co}}{{\text{s}}^2}{\text{A}}}}$                                                                                 $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = 2.{\text{se}}{{\text{c}}^2}{\text{A}}$                                                                                     $\left[ {\because {\text{sec A}} = \dfrac{1}{{{\text{cos A}}}}} \right]$

$ = 2{\mathbf{cose}}{{\mathbf{c}}^2}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{RHS}}$                                 $\left[ {\because {\text{sec}}\left( {90 - {\text{A}}} \right) = {\text{cosec }}{{\text{A}}^{}}} \right]$

$ \Rightarrow {\mathbf{LHS}} = {\mathbf{RHS}}$

Hence proved.

v) $\dfrac{1}{{1\; + \;{\mathbf{sin}}\;\left( {{{90}^ \circ } - {\text{A}}} \right)}} + \dfrac{1}{{1\; - \;{\mathbf{sin}}\;\left( {{{90}^ \circ } - {\text{A}}} \right)}} = 2{\mathbf{se}}{{\mathbf{c}}^2}\left( {{{90}^ \circ } - {\text{A}}} \right)$

Ans:  Taking LHS -

$ = \dfrac{1}{{1\; + \;{\mathbf{sin}}\;\left( {{{90}^ \circ } - {\text{A}}} \right)}} + \dfrac{1}{{1\; - \;{\mathbf{sin}}\;\left( {{{90}^ \circ } - {\text{A}}} \right)}}$

$ = \dfrac{{1\; - \;{\mathbf{cos}}\;{\mathbf{A}} + \left( {1\; + \;{\mathbf{cos}}{\text{A}}} \right)}}{{\left( {1{\text{}} + {\text{cos A}}} \right)\left( {1\; - \;{\mathbf{co}}{\text{sA}}} \right)}}$                                                      $\left[ {\because {\text{sin}}\left( {90 - {\text{A}}} \right) = {\text{cos}}{{\text{A}}^{}}} \right]$

$ = \dfrac{2}{{{1^2} - {\text{Co}}{{\text{s}}^2}{\text{A}}}}$                                                              $\left[ {\because {\text{}}{{\text{a}}^2} - {{\text{b}}^2} = \left( {{\text{a}} + {\text{b}}} \right)\left( {{\text{a}} - {\text{b}}} \right)} \right]$

$ = \dfrac{2}{{{\text{si}}{{\text{n}}^2}{\text{A}}}}$                                                                                 $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = 2.{\text{cose}}{{\text{c}}^2}{\text{A}}$                                                                             $\left[ {\because {\text{cosec A}} = \dfrac{1}{{{\text{sin A}}}}} \right]$

$ = 2{\mathbf{se}}{{\mathbf{c}}^2}\left( {{{90}^ \circ } - {\text{A}}} \right) = {\text{RHS}}$                                     $\left[ {\because {\text{cosec}}\left( {90 - {\text{A}}} \right) = {\text{sec}}{{\text{A}}^{}}} \right]$

$ \Rightarrow {\mathbf{LHS}} = {\mathbf{RHS}}$

Hence proved.

9. If A and B are complementary angles, prove that:

i) ${\mathbf{cot}}\;{\mathbf{B}} + {\mathbf{cos}}\;{\mathbf{B}} = {\mathbf{sec}}\;{\mathbf{A}}.{\mathbf{cos}}\;{\mathbf{B}}\left( {1 + {\mathbf{sin}}\;{\mathbf{B}}} \right)$

Ans: Since, A and B are complementary angles, ${\text{A}} + {\text{B}} = {90^ \circ }$

Taking LHS -

${\mathbf{LHS}} = {\mathbf{cot}}\;{\mathbf{B}} + {\mathbf{cos}}\;{\mathbf{B}}$

$ = {\mathbf{cot}}\;\left( {90 - {\mathbf{A}}} \right) + {\mathbf{cos}}\;\left( {90 - {\mathbf{A}}} \right)$

$ = {\mathbf{tan}}\;{\mathbf{A}} + {\mathbf{sin}}\;{\mathbf{A}}$                $\left[ {\because {\text{cos}}\left( {90 - {{\theta }}} \right) = {\text{sin}}{{{\theta }}^{}}\;\&\; {\text{}}{\mathbf{cot}}\;\left( {90 - {\mathbf{\theta }}} \right) = {\mathbf{tan}}\;{\mathbf{\theta }}} \right]$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}} + {\mathbf{sin}}\;{\mathbf{A}}$                             $\left[ {\because {\text{tan A}} = \dfrac{{{\text{sin A}}}}{{{\text{cos A}}}}} \right]$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{A}}\;.\;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\;\left( {1\; + \;{\mathbf{cos}}\;{\mathbf{A}}} \right)}}{{{\mathbf{cos}}\;{\mathbf{A}}}}$

$ = {\mathbf{sec}}\;{\mathbf{A}}.\;{\mathbf{sin}}\;{\mathbf{A}}\left( {1 + {\mathbf{cos}}\;{\mathbf{A}}} \right)$                                                         $\left[ {\because {\text{sec A}} = \dfrac{1}{{{\text{cos A}}}}} \right]$

$ = {\mathbf{sec}}\;{\mathbf{A}}.\;{\mathbf{sin}}\;\left( {90 - {\mathbf{B}}} \right)\left[ {1 + {\mathbf{cos}}\;\left( {90 - {\mathbf{B}}} \right)} \right]$

$ = {\mathbf{sec}}\;{\mathbf{A}}.{\mathbf{cos}}\;{\mathbf{B}}\left( {1 + {\mathbf{sin}}\;{\mathbf{B}}} \right) = {\mathbf{RHS}}$        $\left[ {\because {\text{sin}}\left( {90 - {{\theta }}} \right) = {\text{cos}}{{{\theta }}^{}}\;\&\; {\text{}}{\mathbf{cos}}\;\left( {90 - {\mathbf{\theta }}} \right) = {\mathbf{sin}}\;{\mathbf{\theta }}} \right]$

$ \Rightarrow {\mathbf{LHS}} = {\mathbf{RHS}}$

Hence proved.

ii) ${\mathbf{cot}}\;{\mathbf{A}}.\;{\mathbf{cot}}\;{\mathbf{B}} - {\mathbf{sin}}\;{\mathbf{A}}.{\mathbf{cos}}\;{\mathbf{B}} - {\mathbf{cos}}\;{\mathbf{A}}.{\mathbf{sin}}\;{\mathbf{B}} = 0$

Ans: Since, A and B are complementary angles, ${\text{A}} + {\text{B}} = {90^ \circ }$

Taking LHS -

${\mathbf{LHS}} = {\mathbf{cot}}\;{\mathbf{A}}.\;{\mathbf{cot}}\;{\mathbf{B}} - {\mathbf{sin}}\;{\mathbf{A}}.{\mathbf{cos}}\;{\mathbf{B}} - {\mathbf{cos}}\;{\mathbf{A}}.{\mathbf{sin}}\;{\mathbf{B}}$

$ = {\mathbf{cot}}\;{\mathbf{A}}.\;{\mathbf{cot}}\;\left( {90 - {\mathbf{A}}} \right) - {\mathbf{sin}}\;{\mathbf{A}}.{\mathbf{cos}}\;\left( {90 - {\mathbf{A}}} \right) - {\mathbf{cos}}\;{\mathbf{A}}.{\mathbf{sin}}\;\left( {90 - {\mathbf{A}}} \right)$

$ = {\mathbf{cot}}\;{\mathbf{A}}.\;{\mathbf{tan}}\;{\mathbf{A}} - {\mathbf{sin}}\;{\mathbf{A}}.{\mathbf{sin}}\;{\mathbf{A}} - {\mathbf{cos}}\;{\mathbf{A}}.{\mathbf{cos}}\;{\mathbf{A}}$

$\left[ {\because {\text{cos}}\left( {90 - {{\theta }}} \right) = {{sin\theta}}\;\;\& {\text{}}{\mathbf{cot}}\;\left( {90 - {\mathbf{\theta }}} \right) = {\mathbf{tan}}\;{\mathbf{\theta }}\;\; \& \;{\text{sin}}\left( {90 - {{\theta }}} \right) = {{cos\theta }}} \right]$

$ = 1 - {\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} - {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}$

$ = 1 - \left( {{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}} \right)$

$ = 1 - 1$                                                                               $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = 0 = {\mathbf{RHS}}$

$ \Rightarrow {\mathbf{LHS}} = {\mathbf{RHS}}$

Hence proved.

iii) ${\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}} + {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{B}} = {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}.{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{B}}$

Ans: Since, A and B are complementary angles, ${\text{A}} + {\text{B}} = {90^ \circ }$

Taking LHS -

${\mathbf{LHS}} = {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}} + {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{B}}$

$ = {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}} + {\mathbf{cose}}{{\mathbf{c}}^2}\left( {90 - {\mathbf{A}}} \right)$

$ = {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}} + {\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}$ $\left[{\because{\text{cos}}{\mathbf{ec}}\left( {90 - {{\theta }}} \right) = {\text{sec}}{{{\theta }}^{}}} \right]$

$ = \dfrac{1}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}} + \dfrac{1}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$                                         $\left[ {\because {\text{sec A}} = \dfrac{1}{{{\text{cos A}}}}{\text{}}\;\&\; {\text{cosec A}} = \dfrac{1}{{{\text{sin A}}}}{\text{}}} \right]$

$ = \dfrac{{\left( {{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}} \right)}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\;.\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$

$=\dfrac{1}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\;.\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$    $\left[{\because{\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$={\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}.{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}$                                      $\left[ {\because {\text{sec A}} = \dfrac{1}{{{\text{coS A}}}}{\text{}}\;\&\; {\text{cosec A}} = \dfrac{1}{{{\text{sin A}}}}{\text{}}} \right]$

$ = {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}.{\mathbf{se}}{{\mathbf{c}}^2}\left( {90 - {\mathbf{B}}} \right)$

$ = {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}.{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{B}} = {\mathbf{RHS}}$                                 $\left[ {\because {\text{s}}{\mathbf{ec}}\left( {90 - {{\theta }}} \right) = {\text{cosec}}{{{\theta }}^{}}} \right]$

$ \Rightarrow {\mathbf{LHS}} = {\mathbf{RHS}}$

Hence proved.

iv) $\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{B}}}}{{{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{B}}}} + \dfrac{{{\mathbf{cos}}\;{\mathbf{B}}\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{B}}\; + \;{\mathbf{cos}}\;{\mathbf{A}}}} = \dfrac{2}{{2{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;1}}$

Ans: Since, A and B are complementary angles, ${\text{A}} + {\text{B}} = {90^ \circ }$

Taking LHS -

${\mathbf{LHS}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{B}}}}{{{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{B}}}} + \dfrac{{{\mathbf{cos}}\;{\mathbf{B}}\; - \;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{B}}\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{B}}}}{{{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{B}}}} + \dfrac{{{\mathbf{cos}}\;\left( {90 - {\mathbf{A}}} \right)\; - \;{\mathbf{cos}}\;\left( {90 - {\mathbf{B}}} \right)}}{{{\mathbf{cos}}\;\left( {90 - {\mathbf{A}}} \right)\; + \;{\mathbf{cos}}\;\left( {90 - {\mathbf{B}}} \right)}}$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{B}}}}{{{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{B}}}} + \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{B}}}}{{{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{B}}}}$                                           $\left[ {\because {\text{cos}}\left( {90 - {{\theta }}} \right) = {\text{sin}}{{{\theta }}^{}}} \right]$

$ = \dfrac{{{{\left( {{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{B}}} \right)}^2}\; + \;{{\left( {{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{B}}} \right)}^2}}}{{\left( {{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{B}}} \right)\left( {{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{sin}}\;{\mathbf{B}}} \right)}}$

$ = \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}\; + \;2\;{\mathbf{sin}}\;{\mathbf{A}}\;.\;{\mathbf{sin}}\;{\mathbf{B}}\; + \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}\; - \;2\;{\mathbf{sin}}\;{\mathbf{A}}\;.\;{\mathbf{sin}}\;{\mathbf{B}}}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}}$

$\left[ {\because {\text{\;\;}}{{\left( {{\mathbf{a}} \pm {\mathbf{b}}} \right)}^2} = {{\mathbf{a}}^2} + {{\mathbf{b}}^2} \pm 2{\mathbf{ab}}\;\;\& \;\left( {{\mathbf{a}} + {\mathbf{b}}} \right)\left( {{\mathbf{a}} - {\mathbf{b}}} \right) = {{\mathbf{a}}^2} - {{\mathbf{b}}^2}} \right]$

$ = \dfrac{{2\left( {{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}} \right)}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}}$

$ = \dfrac{{2\left[ {{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{si}}{{\mathbf{n}}^2}\left( {90 - {\mathbf{A}}} \right)} \right]}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}\left( {90 - {\mathbf{A}}} \right)}}$

$ = \dfrac{{2\left[ {{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}} \right]}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$                                                             $\left[ {\because {\text{s}}{\mathbf{in}}\left( {90 - {{\theta }}} \right) = {\text{cos}}{{{\theta }}^{}}} \right]$

$ = \dfrac{2}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;\left( {1\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}} \right)}}$                                                                $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{2}{{2{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;1}} = {\mathbf{RHS}}$                            

$ \Rightarrow {\mathbf{LHS}} = {\mathbf{RHS}}$

Hence proved.

10. Prove that:

i) $\dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{cos}}\;{\mathbf{A}}}} - \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{A}}}} = \dfrac{{2\;{\mathbf{cos}}\;{\mathbf{A}}}}{{2\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;1}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{cos}}\;{\mathbf{A}}}} - \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{A}}\; - \;\left( {{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{cos}}\;{\mathbf{A}}} \right)}}{{\left( {{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{cos}}\;{\mathbf{A}}} \right)\left( {{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{A}}} \right)}}$

$ = \dfrac{{2\;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$                                                          $\left[ {\because {\text{}}\left( {{\mathbf{a}} + {\mathbf{b}}} \right)\left( {{\mathbf{a}} - {\mathbf{b}}} \right) = {{\mathbf{a}}^2} - {{\mathbf{b}}^2}} \right]$

$ = \dfrac{{2\;{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;\left( {1\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}} \right)}}$                                                                $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{{2\;{\mathbf{cos}}\;{\mathbf{A}}}}{{2\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; - \;1}} = {\mathbf{RHS}}$

$ \Rightarrow {\mathbf{LHS}} = {\mathbf{RHS}}$

Hence proved.

ii) $\dfrac{{{\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}}}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; - \;1}} - 1 = {\mathbf{cosec}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{{\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}}}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; - \;1}} - 1$

$ = \dfrac{{{\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}}\; - \;\left( {{\mathbf{cosec}}\;{\mathbf{A}}\; - \;1} \right)}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; - \;1}}$

$ = \dfrac{{{\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}}\; - \;{\mathbf{cosec}}\;{\mathbf{A}}\; + \;1}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; - \;1}}$                                           

$ = \dfrac{{{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}\; - \;{\mathbf{cosec}}\;{\mathbf{A}}}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; - \;1}}$                                                              $\left[ {\because {\text{}}{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}} = {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} + 1} \right]$

$ = {\mathbf{cosec}}\;{\mathbf{A}}.\dfrac{{\left( {{\mathbf{cosec}}\;{\mathbf{A}}\; - \;1} \right)}}{{{\mathbf{cosec}}\;{\mathbf{A}}\; - \;1}}$

$ = {\mathbf{cosec}}\;{\mathbf{A}} = {\mathbf{RHS}}$

$ \Rightarrow {\mathbf{LHS}} = {\mathbf{RHS}}$

Hence proved.

iii) $\dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}} = {\mathbf{sec}}\;{\mathbf{A}} - {\mathbf{tan}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}} \times \dfrac{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}\left( {1\; - \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}}{{1\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}$                                                       $\left[ {\because {\text{}}\left( {{\mathbf{a}} + {\mathbf{b}}} \right)\left( {{\mathbf{a}} - {\mathbf{b}}} \right) = {{\mathbf{a}}^2} - {{\mathbf{b}}^2}} \right]$                   

$ = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}\left( {1\; - \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}$                                                                    $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}} - \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}$

$ = {\mathbf{sec}}\;{\mathbf{A}} - {\mathbf{tan}}\;{\mathbf{A}} = {\mathbf{RHS}}$

$ \Rightarrow {\mathbf{LHS}} = {\mathbf{RHS}}$

Hence proved.

iv) ${\mathbf{cos}}\;{\mathbf{A}}\left( {1 + {\mathbf{cot}}\;{\mathbf{A}}} \right) + {\mathbf{sin}}\;{\mathbf{A}}\left( {1 + {\mathbf{tan}}\;{\mathbf{A}}} \right) = {\mathbf{sec}}\;{\mathbf{A}} + {\mathbf{cosec}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = {\mathbf{cos}}\;{\mathbf{A}}\left( {1 + {\mathbf{cot}}\;{\mathbf{A}}} \right) + {\mathbf{sin}}\;{\mathbf{A}}\left( {1 + {\mathbf{tan}}\;{\mathbf{A}}} \right)$

$ = {\mathbf{cos}}\;{\mathbf{A}} + {\mathbf{cos}}\;{\mathbf{A}}.{\mathbf{cot}}\;{\mathbf{A}} + {\mathbf{sin}}\;{\mathbf{A}} + {\mathbf{sin}}\;{\mathbf{A}}.{\mathbf{tan}}\;{\mathbf{A}}$

$ = {\mathbf{cos}}\;{\mathbf{A}} + {\mathbf{cos}}\;{\mathbf{A}}.\dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}} + {\mathbf{sin}}\;{\mathbf{A}} + {\mathbf{sin}}\;{\mathbf{A}}.\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}$    $\left[ {\because {\text{tan A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}\;\& \;{\mathbf{cot}}\;{\mathbf{A}} = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = {\mathbf{cos}}\;{\mathbf{A}} + \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}} + {\mathbf{sin}}\;{\mathbf{A}} + \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}$. 

$ = \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; + \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; + \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}$

$ = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}} + \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}$                                $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = {\mathbf{cosec}}\;{\mathbf{A}} + {\mathbf{sec}}\;{\mathbf{A}} = {\mathbf{RHS}}$                          $\left[ {\because {\text{sec A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}\;\& \;{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

Hence proved.

v) $\left( {{\mathbf{sin}}\;{\mathbf{A}} - {\mathbf{cos}}\;{\mathbf{A}}} \right)\left( {1 + {\mathbf{tan}}\;{\mathbf{A}} + {\mathbf{cot}}\;{\mathbf{A}}} \right) = \dfrac{{{\mathbf{sec}}\;{\mathbf{A}}}}{{{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}}} - \dfrac{{{\mathbf{cosec}}\;{\mathbf{A}}}}{{{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \left( {{\mathbf{sin}}\;{\mathbf{A}} - {\mathbf{cos}}\;{\mathbf{A}}} \right)\left( {1 + {\mathbf{tan}}\;{\mathbf{A}} + {\mathbf{cot}}\;{\mathbf{A}}} \right)$

$ = \left( {{\mathbf{sin}}\;{\mathbf{A}} - {\mathbf{cos}}\;{\mathbf{A}}} \right)\left( {1 + \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right)$          $\left[ {\because {\text{tan A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}\;\& \;{\mathbf{cot}}\;{\mathbf{A}} = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = {\mathbf{sin}}\;{\mathbf{A}} + {\mathbf{sin}}\;{\mathbf{A}}.\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}} + {\mathbf{sin}}\;{\mathbf{A}}.\dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}} - {\mathbf{cos}}\;{\mathbf{A}} - {\mathbf{cos}}\;{\mathbf{A}}.\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}} - {\mathbf{cos}}\;{\mathbf{A}}.\dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}$ 

$ = {\mathbf{sin}}\;{\mathbf{A}} + {\mathbf{sin}}\;{\mathbf{A}}.\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}} + {\mathbf{cos}}\;{\mathbf{A}} - {\mathbf{cos}}\;{\mathbf{A}} - {\mathbf{sin}}\;{\mathbf{A}} - {\mathbf{cos}}\;{\mathbf{A}}.\dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}} - \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}$

$ = \dfrac{{{\mathbf{sec}}\;{\mathbf{A}}}}{{{\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}}} - \dfrac{{{\mathbf{cosec}}\;{\mathbf{A}}}}{{{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}}} = {\mathbf{RHS}}$                          $\left[ {\because {\text{sec A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}\;\& \;{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ \Rightarrow {\mathbf{LHS}} = {\mathbf{RHS}}$

Hence proved.

vi) $\sqrt {{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}} + {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}}  = {\mathbf{tan}}\;{\mathbf{A}} + {\mathbf{cot}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \sqrt {{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}} + {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}}} $

$ = \sqrt {\dfrac{1}{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}} + \dfrac{1}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}} $                                         $\left[ {\because {\text{sec A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}\;\& \;{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = \sqrt {\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\;.\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}} $ 

$ = \sqrt {\dfrac{1}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\;.\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}} $                                                                    $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}\;.\;{\mathbf{cos}}\;{\mathbf{A}}}}$

Taking RHS -

${\mathbf{RHS}} = {\mathbf{tan}}\;{\mathbf{A}} + {\mathbf{cot}}\;{\mathbf{A}}$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}$                                                 $\left[ {\because {\text{tan A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}\;\& \;{\mathbf{cot}}\;{\mathbf{A}} = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}\;.\;{\mathbf{cos}}\;{\mathbf{A}}}}$ 

$ = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}\;.\;{\mathbf{cos}}\;{\mathbf{A}}}}$                                                                         $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ \Rightarrow {\mathbf{LHS}} = {\mathbf{RHS}}$

Hence proved.

vii) $\left( {{\mathbf{sin}}\;{\mathbf{A}} + {\mathbf{cos}}\;{\mathbf{A}}} \right)\left( {{\mathbf{sec}}\;{\mathbf{A}} + {\mathbf{cosec}}\;{\mathbf{A}}} \right) = 2 + {\mathbf{sec}}\;{\mathbf{A}}.\;{\mathbf{cosec}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\mathbf{LHS}} = \left( {{\mathbf{sin}}\;{\mathbf{A}} + {\mathbf{cos}}\;{\mathbf{A}}} \right)\left( {{\mathbf{sec}}\;{\mathbf{A}} + {\mathbf{cosec}}\;{\mathbf{A}}} \right)$

$ = \left( {{\mathbf{sin}}\;{\mathbf{A}} + {\mathbf{cos}}\;{\mathbf{A}}} \right)\left( {\dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}} + \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right)$               $\left[ {\because {\text{sec A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}\;\& \;{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}} + 1 + 1 + \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}$ 

$ = 2 + \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}$

$ = 2 + \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}\;.\;{\mathbf{cos}}\;{\mathbf{A}}}}$

$ = 2 + \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}\;.\;{\mathbf{cos}}\;{\mathbf{A}}}}$                                                                 $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = 2 + {\mathbf{sec}}\;{\mathbf{A}}.\;{\mathbf{cosec}}\;{\mathbf{A}} = {\mathbf{RHS}}$                     $\left[ {\because {\text{sec A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}\;\& \;{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ \Rightarrow {\mathbf{LHS}} = {\mathbf{RHS}}$

Hence proved.

viii) $\left( {{\mathbf{tan}}\;{\mathbf{A}} + {\mathbf{cot}}\;{\mathbf{A}}} \right)\left( {{\mathbf{cosec}}\;{\mathbf{A}} - {\mathbf{sin}}\;{\mathbf{A}}} \right)\left( {{\mathbf{sec}}\;{\mathbf{A}} - {\mathbf{cos}}\;{\mathbf{A}}} \right) = 1$

Ans: Taking LHS -

${\mathbf{LHS}} = \left( {{\mathbf{tan}}\;{\mathbf{A}} + {\mathbf{cot}}\;{\mathbf{A}}} \right)\left( {{\mathbf{cosec}}\;{\mathbf{A}} - {\mathbf{sin}}\;{\mathbf{A}}} \right)\left( {{\mathbf{sec}}\;{\mathbf{A}} - {\mathbf{cos}}\;{\mathbf{A}}} \right)$

$ = \left( {\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right)\left( {\dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}} - {\mathbf{sin}}\;{\mathbf{A}}} \right)\left( {\dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}} - {\mathbf{cos}}\;{\mathbf{A}}} \right)$

$\left[ {\because {\text{\;sec\;A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}\;\& \;{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}\;\& \;{\text{tan\;A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}\;\& \;{\mathbf{cot}}\;{\mathbf{A}} = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = \left( {\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\; + \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}\;.\;{\mathbf{cos}}\;{\mathbf{A}}}}} \right)\left( {\dfrac{{1\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right)\left( {\dfrac{{1\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right)$ 

$ = \left( {\dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}\;.\;{\mathbf{cos}}\;{\mathbf{A}}}}} \right)\left( {\dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right)\left( {\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right)$                                              $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = 1 = {\mathbf{RHS}}$

$ \Rightarrow {\mathbf{LHS}} = {\mathbf{RHS}}$

Hence proved.

ix) ${\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} - {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{B}} = \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\;.\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}} = {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}} - {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{B}}$

Ans: : Taking LHS -

${\mathbf{LHS}} = {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{A}} - {\mathbf{co}}{{\mathbf{t}}^2}{\mathbf{B}}$

$=\dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}} - \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}}$                                 $\left[{\because{\text{}}{\mathbf{cot}}\;{\mathbf{A}} = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$=\dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}.{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\;-\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}.\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\;.\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}}$ 

$=\dfrac{{\left({1\;\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}}\right).\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; - \;\left( {1\; -\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}} \right).\;{\mathbf{co}}{{\mathbf{s}}=\dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\;-\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}.\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}\;-\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}\;+\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}.\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\;.\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}}$^2}{\mathbf{B}}}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\;.\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}}$                                      $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\;.\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}} = {\mathbf{MS}}$

Taking MS -

${\mathbf{MS}} = \dfrac{{{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}}\; - \;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{B}}}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\;.\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}}$

$ = \dfrac{{1\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} - \;\left( {1\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}} \right)}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\;.\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}}$                                                            $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\;.\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}}$

$ = \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\;.\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}} - \dfrac{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}\;.\;{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}}$

$ = \dfrac{1}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}}} - \dfrac{1}{{{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{B}}}}$

$ = {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{A}} - {\mathbf{cose}}{{\mathbf{c}}^2}{\mathbf{B}} = {\mathbf{RHS}}$

$ \Rightarrow {\mathbf{LHS}} = {\mathbf{MS}} = {\mathbf{RHS}}$

Hence proved.

x) $\dfrac{{{\mathbf{cot}}\;{\mathbf{A}}\; - \;1}}{{2\; - \;{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}}} = \dfrac{{{\mathbf{cot}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{tan}}\;{\mathbf{A}}}}$

Ans: : Taking LHS -

${\mathbf{LHS}} = \dfrac{{{\mathbf{cot}}\;{\mathbf{A}}\; - \;1}}{{2\; - \;{\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}}}}$

$ = \dfrac{{\dfrac{1}{{{\mathbf{tan}}\;{\mathbf{A}}}}\; - \;1}}{{2\; - \;\left( {1\; + \;{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}} \right)}}$                                        $\left[ {\because {\text{}}1 + {\text{ta}}{{\text{n}}^2}{\text{A}} = {\text{se}}{{\text{c}}^2}{\text{A}}\;\&\; {\text{}}{\mathbf{cot}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{tan}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{\dfrac{{1 - {\mathbf{tan}}\;{\mathbf{A}}}}{{{\mathbf{tan}}\;{\mathbf{A}}}}}}{{\left( {1\; - \;{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}} \right)}}$ 

$ = \dfrac{{\left( {1\; - \;{\mathbf{tan}}\;{\mathbf{A}}} \right)\;{\mathbf{cot}}\;{\mathbf{A}}}}{{\left( {1\; - \;{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}} \right)}}$                                                                              $\left[ {\because {\text{}}{\mathbf{cot}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{tan}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{\left( {1\; - \;{\mathbf{tan}}\;{\mathbf{A}}} \right)\;{\mathbf{cot}}\;{\mathbf{A}}}}{{\left( {1\; - \;{\mathbf{tan}}\;{\mathbf{A}}} \right)\;\left( {1\; + \;{\mathbf{tan}}\;{\mathbf{A}}} \right)}}$                                               $\left[ {\because {\text{}}\left( {{\mathbf{a}} + {\mathbf{b}}} \right)\left( {{\mathbf{a}} - {\mathbf{b}}} \right) = {{\mathbf{a}}^2} - {{\mathbf{b}}^2}} \right]$

$ = \dfrac{{{\mathbf{cot}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{tan}}\;{\mathbf{A}}}} = {\mathbf{RHS}}$

$ \Rightarrow {\mathbf{LHS}} = {\mathbf{RHS}}$

Hence proved.

11. If $4\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} - 3 = 0$ and ${0^ \circ } \leqslant {\mathbf{A}} \leqslant {90^ \circ }$, then prove that:

i) ${\mathbf{sin}}\;3{\mathbf{A}} = 3\;{\mathbf{sin}}\;{\mathbf{A}} - 4\;{\mathbf{si}}{{\mathbf{n}}^3}{\mathbf{A}}$

Ans: Given $4\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} - 3 = 0$

$ \Rightarrow 4\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 3$

$ \Rightarrow {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = \dfrac{3}{4}$

$ \Rightarrow {\mathbf{cos}}\;{\mathbf{A}} = \dfrac{{\sqrt 3 }}{2}$

We know that: ${\text{cos}}{30^ \circ } = \dfrac{{\sqrt 3 }}{2}$

$ \Rightarrow {\text{cosA}} = {\text{cos}}{30^ \circ }$

$ \Rightarrow {\text{A}} = {30^ \circ }$

Taking LHS - 

${\text{LHS}} = {\text{sin}}3{\text{A}}$

$ = {\text{sin}}{90^ \circ }$

$ = 1$

Taking RHS - 

${\text{RHS}} = 3\;{\mathbf{sin}}\;{\mathbf{A}} - 4\;{\mathbf{si}}{{\mathbf{n}}^3}{\mathbf{A}}$

$ = 3{\text{sin}}{30^ \circ } - 4{\text{si}}{{\text{n}}^3}{30^ \circ }$

$ = 3 \times \dfrac{1}{2} - 4{\left( {\dfrac{1}{2}} \right)^3}$

$ = \dfrac{3}{2} - \dfrac{4}{8}$

$ = 1$

$ \Rightarrow {\mathbf{LHS}} = {\mathbf{RHS}}$

Hence proved.

ii) ${\mathbf{cos}}\;3{\mathbf{A}} = 4{\mathbf{co}}{{\mathbf{s}}^3}{\mathbf{A}} - 3\;{\mathbf{cos}}\;{\mathbf{A}}$

Ans: Given $4\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} - 3 = 0$

$ \Rightarrow 4\;{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 3$

$ \Rightarrow {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = \dfrac{3}{4}$

$ \Rightarrow {\mathbf{cos}}\;{\mathbf{A}} = \dfrac{{\sqrt 3 }}{2}$

We know that: ${\text{cos }}{30^ \circ } = \dfrac{{\sqrt 3 }}{2}$

$ \Rightarrow {\text{cos A}} = {\text{cos}}{30^ \circ }$

$ \Rightarrow {\text{A}} = {30^ \circ }$

Taking LHS - 

${\text{LHS}} = {\text{cos}}3{\text{A}}$

$ = {\text{cos}}{90^ \circ }$

$ = 0$

Taking RHS - 

${\text{RHS}} = 4\;{\mathbf{co}}{{\mathbf{s}}^3}{\mathbf{A}} - 3\;{\mathbf{cos}}\;{\mathbf{A}}$

$ = 4{\text{co}}{{\text{s}}^3}{30^ \circ } - 3{\text{cos}}{30^ \circ }$

$ = 4{\left( {\dfrac{{\sqrt 3 }}{2}} \right)^3} - 3 \times \dfrac{{\sqrt 3 }}{2}$

$ = \dfrac{{3\sqrt 3 }}{2} - 3 \times \dfrac{{\sqrt 3 }}{2}$

$ = 0$

$ \Rightarrow {\mathbf{LHS}} = {\mathbf{RHS}}$

Hence proved.

12. Find A, if ${0^ \circ } \leqslant {\mathbf{A}} \leqslant {90^ \circ }$ and:

i) $2{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} - 1 = 0$

Ans: $2{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} - 1 = 0$

$ \Rightarrow 2{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1$

$ \Rightarrow {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = \dfrac{1}{2}$

$ \Rightarrow {\mathbf{cos}}\;{\mathbf{A}} =  \pm \dfrac{1}{{\sqrt 2 }}$

Here given that ${0^ \circ } \leqslant {\text{A}} \leqslant {90^ \circ }$, So in this interval cos A is positive. So we will neglect negative values.

$ \Rightarrow {\mathbf{cos}}\;{\mathbf{A}} = \dfrac{1}{{\sqrt 2 }}$

We know that ${\text{cos}}{45^ \circ } = \dfrac{1}{{\sqrt 2 }}$

$ \Rightarrow {\text{cos A}} = {\text{cos }}{45^ \circ }$

$ \Rightarrow {\mathbf{A}} = {45^ \circ }$

ii) ${\mathbf{sin}}\;3{\mathbf{A}} - 1 = 0$

$ \Rightarrow {\mathbf{sin}}\;3{\mathbf{A}} = 1$

We know that ${\text{sin}}{90^ \circ } = 1$

$ \Rightarrow {\text{sin}}3{\text{A}} = {\text{sin}}{90^ \circ }$

$ \Rightarrow 3{\mathbf{A}} = {90^ \circ }$

$ \Rightarrow {\mathbf{A}} = {30^ \circ }$

iii) $4{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} - 3 = 0$

Ans: $4{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} - 3 = 0$

$ \Rightarrow 4{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} = 3$

$ \Rightarrow {\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} = \dfrac{3}{4}$

$ \Rightarrow {\mathbf{sin}}\;{\mathbf{A}} =  \pm \dfrac{{\sqrt 3 }}{2}$

Here given that ${0^ \circ } \leqslant {\text{A}} \leqslant {90^ \circ }$, So in this interval sin A is positive. So we will neglect negative values.

$ \Rightarrow {\mathbf{sin}}\;{\mathbf{A}} = \dfrac{{\sqrt 3 }}{2}$

We know that ${\text{sin}}{60^ \circ } = \dfrac{{\sqrt 3 }}{2}$

$ \Rightarrow {\mathbf{sin}}\;{\mathbf{A}} = {\text{sin}}{60^ \circ }$

$ \Rightarrow {\mathbf{A}} = {60^ \circ }$

iv) ${\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} - {\mathbf{cos}}\;{\mathbf{A}} = 0$

Ans: ${\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} - {\mathbf{cos}}\;{\mathbf{A}} = 0$

$ \Rightarrow {\mathbf{cos}}\;{\mathbf{A}}\left( {{\mathbf{cos}}\;{\mathbf{A}} - 1} \right) = 0$

$ \Rightarrow {\mathbf{cos}}\;{\mathbf{A}} = 0$ Or ${\mathbf{cos}}\;{\mathbf{A}} - 1 = 0$

$ \Rightarrow {\mathbf{cos}}\;{\mathbf{A}} = 0$ Or ${\mathbf{cos}}\;{\mathbf{A}} = 1$

We know that ${\text{cos}}{90^ \circ } = 0{\text{}}\;\&\; {\text{cos}}{0^ \circ } = 1$

$ \Rightarrow {\mathbf{cos}}\;{\mathbf{A}} = {\mathbf{cos}}\;{90^ \circ }$ Or ${\mathbf{cos}}\;{\mathbf{A}} = {\mathbf{cos}}\;{0^ \circ }$

$ \Rightarrow {\text{A}} = {90^ \circ }$ or ${0^ \circ }$

v) $2{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} + {\mathbf{cos}}\;{\mathbf{A}} - 1 = 0$

Ans: $2{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} + {\mathbf{cos}}\;{\mathbf{A}} - 1 = 0$

$ \Rightarrow 2{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} + 2{\mathbf{cos}}\;{\mathbf{A}} - {\mathbf{cos}}\;{\mathbf{A}} - 1 = 0$

$ \Rightarrow 2{\mathbf{cos}}\;{\mathbf{A}}\left( {{\mathbf{cos}}\;{\mathbf{A}} + 1} \right) - 1\left( {{\mathbf{cos}}\;{\mathbf{A}} + 1} \right) = 0$

$ \Rightarrow \left( {2{\mathbf{cos}}\;{\mathbf{A}} - 1} \right)\left( {{\mathbf{cos}}\;{\mathbf{A}} + 1} \right) = 0$

$ \Rightarrow 2{\mathbf{cos}}\;{\mathbf{A}} - 1 = 0$ or ${\mathbf{cos}}\;{\mathbf{A}} + 1 = 0$

$ \Rightarrow {\mathbf{cos}}\;{\mathbf{A}} = \dfrac{1}{2}$ or ${\mathbf{cos}}\;{\mathbf{A}} =  - 1$

We know that ${\text{cos}}{60^ \circ } = \dfrac{1}{2}$

We also know that for no value of A in ${0^ \circ } \leqslant {\text{A}} \leqslant {90^ \circ }$, ${\mathbf{cos}}\;{\mathbf{A}} =  - 1$

$ \Rightarrow {\mathbf{cos}}\;{\mathbf{A}} = {\mathbf{cos}}\;{60^ \circ }$

$ \Rightarrow {\text{A}} = {60^ \circ }$

13. If ${0^ \circ } \leqslant {\text{A}} \leqslant {90^ \circ }$; find A, if:

i) $\dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}} = 4$

Ans: $\dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; - \;{\mathbf{sin}}\;{\mathbf{A}}}} + \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{1\; + \;{\mathbf{sin}}\;{\mathbf{A}}}} = 4$

$ \Rightarrow \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}\left( {1\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)\; + \;{\mathbf{cos}}\;{\mathbf{A}}\left( {1\; - \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}}{{\left( {1\; + \;{\mathbf{sin}}\;{\mathbf{A}}} \right)\left( {1\; - \;{\mathbf{sin}}\;{\mathbf{A}}} \right)}} = 4$

$ \Rightarrow \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{A}}.{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{cos}}\;{\mathbf{A}}\; - \;{\mathbf{cos}}\;{\mathbf{A}}.{\mathbf{sin}}\;{\mathbf{A}}}}{{\left( {{1^2}\; - \;{\mathbf{si}}{{\mathbf{n}}^2}\;{\mathbf{A}}} \right)}} = 4$

$ \Rightarrow \dfrac{{2{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{co}}{{\mathbf{s}}^2}\;{\mathbf{A}}}} = 4$

$ \Rightarrow \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}} = 2$

$ \Rightarrow {\mathbf{cos}}\;{\mathbf{A}} = \dfrac{1}{2}$

We know that ${\text{cos}}{60^ \circ } = \dfrac{1}{2}$

$ \Rightarrow {\mathbf{cos}}\;{\mathbf{A}} = {\mathbf{cos}}\;{60^ \circ }$

$ \Rightarrow {\text{A}} = {60^ \circ }$

ii) $\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{sec}}\;{\mathbf{A}}\; - \;1}} + \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{sec}}\;{\mathbf{A}}\; + \;1}} = 2$

Ans: $\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{sec}}\;{\mathbf{A}}\; - \;1}} + \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{sec}}\;{\mathbf{A}}\; + \;1}} = 2$

$ \Rightarrow \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\left( {1\; + \;{\mathbf{sec}}\;{\mathbf{A}}} \right)\; + \;{\mathbf{sin}}\;{\mathbf{A}}\left( {{\mathbf{sec}}\;{\mathbf{A}}\; - \;1} \right)}}{{\left( {1\; + \;{\mathbf{sec}}\;{\mathbf{A}}} \right)\left( {{\mathbf{sec}}\;{\mathbf{A}}\; - \;1} \right)}} = 2$

$ \Rightarrow \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{sec}}\;{\mathbf{A}}.{\mathbf{sin}}\;{\mathbf{A}}\; - \;{\mathbf{sin}}\;{\mathbf{A}}\; + \;{\mathbf{sec}}\;{\mathbf{A}}.{\mathbf{sin}}\;{\mathbf{A}}}}{{\left( {{\mathbf{se}}{{\mathbf{c}}^2}\;{\mathbf{A}}\; - \;{1^2}} \right)}} = 2$

$ \Rightarrow \dfrac{{2{\mathbf{sec}}\;{\mathbf{A}}.\;{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{ta}}{{\mathbf{n}}^2}\;{\mathbf{A}}}} = 2$

$ \Rightarrow \dfrac{{\dfrac{{2\;{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}}}{{{\mathbf{ta}}{{\mathbf{n}}^2}\;{\mathbf{A}}}} = 2$

$ \Rightarrow \dfrac{{{\mathbf{tan}}\;{\mathbf{A}}}}{{{\mathbf{ta}}{{\mathbf{n}}^2}{\mathbf{A}}}} = 1$

$ \Rightarrow {\mathbf{tan}}\;{\mathbf{A}} = 1$

We know that ${\text{tan}}{45^ \circ } = 1$

$ \Rightarrow {\mathbf{tan}}\;{\mathbf{A}} = {\mathbf{tan}}\;{45^ \circ }$

$ \Rightarrow {\text{A}} = {45^ \circ }$

14. Prove that: 

$\left( {{\mathbf{cosec}}\;{\mathbf{A}} - {\mathbf{sin}}\;{\mathbf{A}}} \right)\left( {{\mathbf{sec}}\;{\mathbf{A}} - {\mathbf{cos}}\;{\mathbf{A}}} \right){\mathbf{se}}{{\mathbf{c}}^2}{\mathbf{A}} = {\mathbf{tan}}\;{\mathbf{A}}$

Ans: Taking LHS -

${\text{LHS}} = \left( {{\text{cosec A}} - {\text{sin A}}} \right)\left( {{\text{sec A}} - {\text{cos A}}} \right){\text{se}}{{\text{c}}^2}{\text{A}}$

$ = \left( {\dfrac{1}{{{\text{sin A}}}} - {\text{sin A}}} \right)\left( {\dfrac{1}{{{\text{cos A}}}} - {\text{cos A}}} \right){\text{se}}{{\text{c}}^2}{\text{A}}$

 $\left[ {\because {\text{sec A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}\;\& \;{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = \left( {\dfrac{{1{\text{}} - {\text{si}}{{\text{n}}^2}{\text{A}}}}{{{\text{sin A}}}}} \right)\left( {\dfrac{{1{\text{}} - {\text{co}}{{\text{s}}^2}{\text{A}}}}{{{\text{cos A}}}}} \right){\text{se}}{{\text{c}}^2}{\text{A}}$

$ = \left( {\dfrac{{{\text{co}}{{\text{s}}^2}{\text{A}}}}{{{\text{sin A}}}}} \right)\left( {\dfrac{{{\text{si}}{{\text{n}}^2}{\text{A}}}}{{{\text{cos A}}}}} \right){\text{se}}{{\text{c}}^2}{\text{A}}$                                    $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}+{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$=\left(\text{cos A}\right){\text{sin A}}.\dfrac{1}{cos^{2}A}$                    $\left[{\because{\text{sec A}}=\dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$

$ = \dfrac{{{\text{sin A}}}}{{{\text{cos A}}}}$

$ = {\text{tan A}} = {\text{RHS}}$                                           $\left[ {\because {\text{tan A}}=\dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}} \right]$

$ \Rightarrow {\mathbf{LHS}} = {\mathbf{RHS}}$

Hence proved.

15. Prove the identity $\left( {{\mathbf{sin}}\;{\mathbf{\theta }} + {\mathbf{cos}}\;{\mathbf{\theta }}} \right)\left( {{\mathbf{tan}}\;{\mathbf{\theta }} + {\mathbf{cot}}\;{\mathbf{\theta }}} \right) = {\mathbf{sec}}\;{\mathbf{\theta }} + {\mathbf{cosec}}\;{\mathbf{\theta }}$.

Ans: Taking LHS -

${\text{LHS}} = \left( {{{sin\;\theta }} + {{cos\;\theta }}} \right)\left( {{{tan\;\theta }} + {{cot\;\theta }}} \right)$

$ = \left( {{{sin\;\theta }} + {{cos\;\theta }}} \right)\left( {\dfrac{{{\mathbf{sin}}\;{\mathbf{\theta }}}}{{{\mathbf{cos}}\;{\mathbf{\theta }}}} + \dfrac{{{\mathbf{cos}}\;{\mathbf{\theta }}}}{{{\mathbf{sin}}\;{\mathbf{\theta }}}}} \right)$                  $\left[ {\because {\text{tan A}} = \dfrac{{{\mathbf{sin}}\;{\mathbf{A}}}}{{{\mathbf{cos}}\;{\mathbf{A}}}}\;\& \;{\mathbf{cot}}\;{\mathbf{A}} = \dfrac{{{\mathbf{cos}}\;{\mathbf{A}}}}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ = {{sin\;\theta }}.\dfrac{{{\mathbf{sin}}\;{\mathbf{\theta }}}}{{{\mathbf{cos}}\;{\mathbf{\theta }}}} + {\mathbf{sin}}\;{\mathbf{\theta }}.\dfrac{{{\mathbf{cos}}\;{\mathbf{\theta }}}}{{{\mathbf{sin}}\;{\mathbf{\theta }}}} + {{cos\;\theta }}.\dfrac{{{\mathbf{sin}}\;{\mathbf{\theta }}}}{{{\mathbf{cos}}\;{\mathbf{\theta }}}} + {{cos\;\theta }}.\dfrac{{{\mathbf{cos}}\;{\mathbf{\theta }}}}{{{\mathbf{sin}}\;{\mathbf{\theta }}}}$

$ = \left( {\dfrac{{{\text{si}}{{\text{n}}^2}{{\theta }}}}{{{{cos\;\theta }}}}} \right) + {{cos\;\theta }} + {{sin\;\theta }} + \left( {\dfrac{{{\text{co}}{{\text{s}}^2}{{\theta }}}}{{{{sin\theta }}}}} \right)$

$ = \left( {\dfrac{{{\text{si}}{{\text{n}}^2}{{\theta}} + {\text{co}}{{\text{s}}^2}{{\theta }}}}{{{{cos\;\theta }}}}} \right) + \left( {\dfrac{{{\text{si}}{{\text{n}}^2}{{\theta}} + {\text{co}}{{\text{s}}^2}{{\theta }}}}{{{{sin\theta }}}}} \right)$

$ = \left( {\dfrac{1}{{{{cos\;\theta }}}}} \right) + \left( {\dfrac{1}{{{{sin\;\theta }}}}}\right)$ $\left[{\because{\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}}+{\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = {\mathbf{sec}}\;{\mathbf{\theta }} + {\mathbf{cosec}}\;{\mathbf{\theta }} = {\text{RHS}}$                          $\left[ {\because {\text{sec A}} = \dfrac{1}{{{\mathbf{cos}}\;{\mathbf{A}}}}\;\& \;{\mathbf{cosec}}\;{\mathbf{A}} = \dfrac{1}{{{\mathbf{sin}}\;{\mathbf{A}}}}} \right]$

$ \Rightarrow {\mathbf{LHS}} = {\mathbf{RHS}}$

Hence proved.

16. Evaluate without using trigonometric tables,

${\mathbf{si}}{{\mathbf{n}}^2}{28^ \circ } + {\mathbf{si}}{{\mathbf{n}}^2}{62^ \circ } + {\mathbf{ta}}{{\mathbf{n}}^2}{38^ \circ } - {\mathbf{co}}{{\mathbf{t}}^2}{52^ \circ } + \dfrac{1}{4}{\mathbf{se}}{{\mathbf{c}}^2}{30^ \circ }$

Ans: ${\mathbf{si}}{{\mathbf{n}}^2}{28^ \circ } + {\mathbf{si}}{{\mathbf{n}}^2}{62^ \circ } + {\mathbf{ta}}{{\mathbf{n}}^2}{38^ \circ } - {\mathbf{co}}{{\mathbf{t}}^2}{52^ \circ } + \dfrac{1}{4}{\mathbf{se}}{{\mathbf{c}}^2}{30^ \circ }$

$ = {\mathbf{si}}{{\mathbf{n}}^2}{28^ \circ } + {\mathbf{si}}{{\mathbf{n}}^2}{\left( {90 - 28} \right)^ \circ } + {\mathbf{ta}}{{\mathbf{n}}^2}{38^ \circ } - {\mathbf{co}}{{\mathbf{t}}^2}{\left( {90 - 38} \right)^ \circ } + \dfrac{1}{4}{\mathbf{se}}{{\mathbf{c}}^2}{30^ \circ }$

$ = {\mathbf{si}}{{\mathbf{n}}^2}{28^ \circ } + {\mathbf{co}}{{\mathbf{s}}^2}{28^ \circ } + {\mathbf{ta}}{{\mathbf{n}}^2}{38^ \circ } - {\mathbf{ta}}{{\mathbf{n}}^2}{38^ \circ } + \dfrac{1}{4}{\left( {\dfrac{2}{{\sqrt 3 }}} \right)^2}$

$\left[ {\because {\text{sin}}\left( {90 - {\text{A}}} \right) = {\mathbf{cos}}\;{\mathbf{A}}\;\& \;{\mathbf{cot}}\;\left( {90 - {\mathbf{A}}} \right) = {\mathbf{tan}}\;{\mathbf{A}}\;\& \;{\mathbf{sec}}\;{{30}^ \circ } = \dfrac{2}{{\sqrt 3 }}} \right]$

$ = 1 + 0 + \dfrac{1}{4}\left( {\dfrac{4}{3}} \right)$                                                    $\left[ {\because {\text{}}{\mathbf{si}}{{\mathbf{n}}^2}{\mathbf{A}} + {\mathbf{co}}{{\mathbf{s}}^2}{\mathbf{A}} = 1} \right]$

$ = 1 + \dfrac{1}{3}$

$ = \dfrac{4}{3}$

How Should the Preparation be in Class 10?

Students that are facing difficulties in preparing for the exam and those who do not know how to solve a problem can reach the Vedantu website, for we are providing the students with proper guidance to prepare for the exam. Students that are having difficulties in solving a sum can make use of the Seline publisher’s textbook. This textbook is a best-selling textbook on the ICSE board. Almost all the students are using this textbook for practising purposes. Students might ask what is so good about this textbook? The answer to this question will be your score. This textbook contains all the sums and detailed answers to it. Every solution in this textbook is solved in a step-by-step manner considering step marks have a vital role in mathematics. 

Why is Trigonometry Important in Class 10?

Trigonometric preparation must be intense and more focused. All the formulae must be at the fingertips of the students. Having complicated formulae might be confusing. But the planned preparation can beat the chaos. Practice every day with pen and paper. Do not just see and read the formulae. Write down and make it a cheat sheet that will help you to revise at the neck of the moment before entering the exam hall.

Your preparation must include proper order. Get the updated syllabus from the Vedantu website and understand the pattern. Then improvise your learning by practising more important questions that will help you to get a better idea about the concept. Then move to take up the mock tests. Mock tests are a vital part of the preparation. It gives you a preview of the main examination. You can practice this way to score better marks in the examination. Doubts can be clarified by our Vedantu team. 

Conclusion

All study materials are available for free. Make use of these materials and stand out from other students. Exams are nearing. Start your preparation right now. Get onto the Vedantu website. We are here to help you. All the best for the examination!