Question

$ {\text{If }}\theta = {30^ \circ },{\text{ verify that:}} \\
  \left( {\text{i}} \right){\text{tan2}}\theta {\text{ = }}\dfrac{{2{\text{tan}}\theta }}{{1 - {\text{ta}}{{\text{n}}^2}\theta }} \\
  \left( {{\text{ii}}} \right){\text{tan2}}\theta {\text{ = }}\dfrac{{{\text{4tan}}\theta }}{{1 + {\text{ta}}{{\text{n}}^2}\theta }} \\
  \left( {{\text{iii}}} \right){\text{cos2}}\theta {\text{ = }}\dfrac{{1 - {\text{ta}}{{\text{n}}^2}\theta }}{{1 + {\text{ta}}{{\text{n}}^2}\theta }} \\
  \left( {{\text{iv}}} \right){\text{cos3}}\theta {\text{ = 4co}}{{\text{s}}^3}\theta - 3{\text{cos}}\theta \\$

Answer 1

Hint: In this collection of questions we have to solve Left Hand Side(LHS) and Right Hand Side(RHS) separately for each question. Here we just need to put the value of \[\theta = {30^ \circ }\] in both LHS and RHS. If we get the same value then the LHS = RHS.

Complete step-by-step answer:
\[\left( {\text{i}} \right){\text{tan2}}\theta {\text{ = }}\dfrac{{2{\text{tan}}\theta }}{{1 - {\text{ta}}{{\text{n}}^2}\theta }}\]
On putting$\theta = {30^ \circ }$ in LHS, we get
$
   \Rightarrow {\text{tan2}}\theta \\
   \Rightarrow {\text{tan2(3}}{0^ \circ }) \\
   \Rightarrow \tan {60^ \circ } \\
 $
And we know, $\tan {60^ \circ } = \sqrt 3 $
Then,
$
   \Rightarrow {\text{tan}}{60^ \circ } \\
   \Rightarrow \sqrt 3 {\text{ ---------- eq}}{\text{.1}} \\
$

On putting$\theta = {30^ \circ }$ in RHS, we get
$
   \Rightarrow \dfrac{{2{\text{tan}}\theta }}{{1 - {\text{ta}}{{\text{n}}^2}\theta }} \\
   \Rightarrow \dfrac{{2{\text{tan3}}{0^ \circ }}}{{1 - {\text{ta}}{{\text{n}}^2}{{30}^ \circ }}} \\
$
And we know $\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$
Then,
$
   \Rightarrow \dfrac{{2(\dfrac{1}{{\sqrt 3 }})}}{{1 - {{(\dfrac{1}{{\sqrt 3 }})}^2}}}{\text{ }} \\
   \Rightarrow \sqrt 3 {\text{ -------- eq}}{\text{.2}} \\
$
From eq.1 and eq.2 we observe that
LHS=RHS Hence Proved

\[\left( {{\text{ii}}} \right){\text{tan2}}\theta {\text{ = }}\dfrac{{{\text{4tan}}\theta }}{{1 + {\text{ta}}{{\text{n}}^2}\theta }}\]
On putting$\theta = {30^ \circ }$ in LHS, we get
$
   \Rightarrow {\text{tan2}}\theta \\
   \Rightarrow {\text{tan2(3}}{0^ \circ }) \\
   \Rightarrow {\text{tan}}{60^ \circ } \\
$
And we know, $\tan {60^ \circ } = \sqrt 3 $
Then,
$
   \Rightarrow {\text{tan}}{60^ \circ } \\
   \Rightarrow \sqrt 3 {\text{ --------- eq}}{\text{.3}} \\
 $


On putting$\theta = {30^ \circ }$ in RHS, we get
$
   \Rightarrow \dfrac{{{\text{4tan}}\theta }}{{1 + {\text{ta}}{{\text{n}}^2}\theta }} \\
   \Rightarrow \dfrac{{{\text{4tan3}}{0^ \circ }}}{{1 + {\text{ta}}{{\text{n}}^2}{{30}^ \circ }}} \\
 $
And we know $\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$
Then,

$
   \Rightarrow \dfrac{{4(\dfrac{1}{{\sqrt 3 }})}}{{1 + {{(\dfrac{1}{{\sqrt 3 }})}^2}}} \\
   \Rightarrow \sqrt 3 {\text{ ------- eq}}{\text{.4}} \\
$
From eq.3 and eq.4 we observe that
LHS=RHS Hence Proved

\[\left( {{\text{iii}}} \right){\text{cos2}}\theta {\text{ = }}\dfrac{{1 - {\text{ta}}{{\text{n}}^2}\theta }}{{1 + {\text{ta}}{{\text{n}}^2}\theta }}\]
On putting$\theta = {30^ \circ }$ in LHS, we get
$
   \Rightarrow {\text{cos2}}\theta \\
   \Rightarrow \cos {\text{2(3}}{0^ \circ }) \\
$
And, we $\cos {60^ \circ } = \dfrac{1}{2}$
$
   \Rightarrow \cos {60^ \circ } \\
   \Rightarrow \dfrac{1}{2}{\text{ -------- eq}}{\text{.5}} \\
$
On putting$\theta = {30^ \circ }$ in RHS, we get
$
   \Rightarrow \dfrac{{1 - {\text{ta}}{{\text{n}}^2}\theta }}{{1 + {\text{ta}}{{\text{n}}^2}\theta }} \\
   \Rightarrow \dfrac{{1 - {\text{ta}}{{\text{n}}^2}{{30}^ \circ }}}{{1 + {\text{ta}}{{\text{n}}^2}{{30}^ \circ }}} \\
$
And we know $\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$
Then,

$
   \Rightarrow {\dfrac{{1 - {{(\dfrac{1}{{\sqrt 3 }})}^2}}}{{1 + {{(\dfrac{1}{{\sqrt 3 }})}^2}}}^{}}{\text{ }} \\
   \Rightarrow \dfrac{1}{2}{\text{ ------- eq}}{\text{.6}} \\
 $
From eq.5 and eq.6 we observe that
LHS=RHS Hence Proved

\[\left( {{\text{iv}}} \right){\text{cos3}}\theta {\text{ = 4co}}{{\text{s}}^3}\theta - 3{\text{cos}}\theta \]
On putting$\theta = {30^ \circ }$ in LHS, we get
$
   \Rightarrow {\text{cos2}}\theta \\
   \Rightarrow \cos 3{\text{(3}}{0^ \circ }) \\
 $
And, we know ${\text{cos9}}{0^ \circ } = 0$
Then,
$
   \Rightarrow \cos {90^ \circ } \\
   \Rightarrow 0{\text{ -------eq}}{\text{.7}} \\
 $
On putting$\theta = {30^ \circ }$ in RHS, we get
$
   \Rightarrow 4{\text{co}}{{\text{s}}^3}\theta - {\text{3cos}}\theta \\
   \Rightarrow 4{\text{co}}{{\text{s}}^3}{30^ \circ } - {\text{3cos}}{30^ \circ } \\
$
And, we know ${\text{cos 3}}{0^ \circ } = \dfrac{{\sqrt 3 }}{2}$
Then,
$
   \Rightarrow 4{(\dfrac{{\sqrt 3 }}{2})^3} - 3(\dfrac{{\sqrt 3 }}{2}){\text{ }} \\
   \Rightarrow 0{\text{ -------eq}}{\text{.8}} \\
 $
From eq.7 and eq.8 we observe that
LHS=RHS Hence Proved

Note: Whenever you get this type of question the key concept to solve this question is to evaluate the separate Left Hand Side(LHS) and Right Hand Side(RHS) and check if both are the same then the given question is true otherwise false. And one more thing to remember is to learn the commonly used trigonometric angles values like $\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }},\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$ etc.