Question

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) \[{\left( {cosec\theta - \cot \theta } \right)^2} = \dfrac{{1 - \cos \theta }}{{1 + \cos \theta }}\]
 (ii)\[\dfrac{{\cos A}}{{1 + \sin A}} + \dfrac{{1 + \sin A}}{{\cos A}}\] \[ = \sec A\]
(iii) \[\dfrac{{\tan \theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \tan \theta }} = 1 + \sec \theta \cdot cosec\theta \]
(iv) \[\dfrac{{1 + \sec A}}{{\sec A}} = \dfrac{{{{\sin }^2}A}}{{1 - \cos A}}\]
(v) \[\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \cos ecA + \cot A\]
(vi) \[\sqrt {\dfrac{{1 + \sin A}}{{1 - \sin A}}} = \sec A + \tan A\]
(vii) \[\dfrac{{\sin \theta - 2{{\sin }^3}\theta }}{{2{{\cos }^3}\theta - \cos \theta }} = \tan \theta \]
(viii) \[{\left( {\sin A + \cos ecA} \right)^2} + {\left( {\cos A + \sec A} \right)^2} = 7 + {\tan ^2}A + {\cot ^2}A\]
(ix) \[\left( {cosecA - \sin A} \right)\left( {\sec A - \cos A} \right) = \dfrac{1}{{\tan A + \cot A}}\]
(x) \[\left[ {\dfrac{{1 + {{\tan }^2}A}}{{1 + {{\cot }^2}A}}} \right] = {\left[ {\dfrac{{1 - \tan A}}{{1 - citA}}} \right]^2} = {\tan ^2}A\]

Answer 1

Hint: We will prove in all the parts of the question that the Left Hand Side and the Right Hand Side of the trigonometric equation are equal, for which, we first write down the given equation and start solving either the LHS or the RHS of the equation by taking the help of various trigonometric identities.

Complete step-by-step answer:

(i) \[{\left( {cosec\theta - \cot \theta } \right)^2} = \dfrac{{1 - \cot \theta }}{{1 + \cot \theta }}\]
Taking LHS, we get,
\[{\left( {cosec\theta - \cot \theta } \right)^2}\]……. \[\left( 1 \right)\]
Now, we will make it in terms of \[\sin \theta and\cos \theta \].
As we know that, \[cosec\theta = \dfrac{1}{{\sin \theta }}\]and \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\]
Therefore, equation (1) becomes,
= \[\left( {\dfrac{1}{{\sin \theta }}} \right. - {\left. {\dfrac{{\cos \theta }}{{\sin \theta }}} \right)^2}\]
= \[{\left( {\dfrac{{1 - \cos \theta }}{{\sin \theta }}} \right)^2}\]
= \[\dfrac{{{{\left( {1 - \cos \theta } \right)}^2}}}{{{{\sin }^2}\theta }}\]
We know that, \[{\cos ^2}\theta + {\sin ^2}\theta = 1\]
                      \[ \Rightarrow \]\[{\sin ^2}\theta = 1 - {\cos ^2}\theta \]
Therefore, \[\dfrac{{{{\left( {1 - \cos \theta } \right)}^2}}}{{1 - {{\cos }^2}\theta }}\]
= \[\dfrac{{{{\left( {1 - \cos \theta } \right)}^2}}}{{{1^2} - {{\cos }^2}\theta }}\]
Since, in the denominator , \[{1^2} - {\cos ^2}\theta \] represents an identity, i.e., \[{a^2} - {b^2}\]\[ = \left( {a - b} \right)\left( {a + b} \right)\],then it becomes,
= \[\dfrac{{{{\left( {1 - \cos \theta } \right)}^2}}}{{\left( {1 + \cos \theta } \right)\left( {1 - \cos \theta } \right)}}\]
= \[\dfrac{{\left( {1 - \cos \theta } \right)\left( {1 - \cos \theta } \right)}}{{\left( {1 + \cos \theta } \right)\left( {1 - \cos \theta } \right)}}\]
= \[\dfrac{{1 - \cos \theta }}{{1 + \cos \theta }}\]
=RHS
Hence Proved.

(ii) \[\dfrac{{\cos A}}{{1 + \sin A}} + \dfrac{{1 + \sin A}}{{\cos A}}\]\[ = \sec A\]
Taking LHS, we get
\[\dfrac{{\cos A}}{{1 + \sin A}} + \dfrac{{1 + \sin A}}{{\cos A}}\] , which is already in the terms of sine and cosine
Now, by taking LCM, we get
=\[\dfrac{{{{\cos }^2}A + {{\left( {1 + \sin A} \right)}^2}}}{{\cos A\left( {1 + \sin A} \right)}}\] , where in the numerator, \[{\left( {1 + \sin A} \right)^2}\]represents an identity, i.e., \[{\left( {a + b} \right)^2}\],then it becomes,
= \[\dfrac{{{{\cos }^2}A + 1 + {{\sin }^2}A + 2\sin A}}{{\cos A\left( {1 + \sin A} \right)}}\]
As we know that, \[{\cos ^2}A + {\sin ^2}A = 1\],therefore, we get
= \[\dfrac{{1 + 1 + 2\sin A}}{{\cos A\left( {1 + \sin A} \right)}}\]
= \[\dfrac{{2 + 2\sin A}}{{\cos A\left( {1 + \sin A} \right)}}\]
Now, taking 2 common in the numerator, we get
=\[\dfrac{{2\left( {1 + \sin A} \right)}}{{\cos A\left( {1 + \operatorname{sinA} } \right)}}\]
By cutting equal terms in the numerator and denominator, we get
=\[\dfrac{2}{{\cos A}}\]
=\[2 \cdot \dfrac{1}{{\cos A}}\]
=\[\sec A\]
=RHS
Hence Proved.

(iii) \[\dfrac{{\tan \theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \tan \theta }} = 1 + \sec \theta \cos ec\theta \]
Taking LHS, we get
\[\dfrac{{\tan \theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \tan \theta }}\]

Now, we will make it in terms of \[\sin \theta and\cos \theta \].
We know that, \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\] and \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\], then it becomes,
= \[\dfrac{{\dfrac{{\sin \theta }}{{\cos \theta }}}}{{1 - \dfrac{{\cos \theta }}{{\sin \theta }}}} + \dfrac{{\dfrac{{\cos \theta }}{{\sin \theta }}}}{{1 - \dfrac{{\sin \theta }}{{\cos \theta }}}}\]
Now, by taking LCM in the denominator, we get

=\[\dfrac{{\dfrac{{\sin \theta }}{{\cos \theta }}}}{{\dfrac{{\sin \theta - \cos \theta }}{{\sin \theta }}}} + \dfrac{{\dfrac{{\cos \theta }}{{\sin }}}}{{\dfrac{{\cos \theta - \sin \theta }}{{\cos \theta }}}}\]
By solving, it becomes
= \[\dfrac{{\sin \theta }}{{\cos \theta }} \times \dfrac{{\sin \theta }}{{\sin \theta - \cos \theta }} + \dfrac{{\cos \theta }}{{\sin \theta }} \times \dfrac{{\cos \theta }}{{\cos \theta - \sin \theta }}\]

=\[\dfrac{{{{\sin }^2}\theta }}{{\cos \theta \left( {\sin \theta - \cos \theta } \right)}} + \dfrac{{{{\cos }^2}\theta }}{{\sin \theta \left( {\cos \theta - \sin \theta } \right)}}\]
By taking minus common from the second term,
= \[\dfrac{{{{\sin }^2}\theta }}{{\cos \theta \left( {\sin \theta - \cos \theta } \right)}} - \dfrac{{{{\cos }^2}\theta }}{{\sin \theta \left( {\sin \theta - \cos \theta } \right)}}\]

= \[\dfrac{{{{\sin }^3}\theta - {{\cos }^3}\theta }}{{\cos \theta \cdot \sin \theta \left( {\sin \theta - \cos \theta } \right)}}\]
Now, in the numerator, \[{\sin ^3}\theta - {\cos ^3}\theta \]represents an identity, i.e., \[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\], therefore, it becomes,
= \[\dfrac{{\left( {\sin \theta - \cos \theta } \right)\left( {{{\sin }^2}\theta + \sin \theta \cdot \cos \theta + {{\cos }^2}\theta } \right)}}{{\cos \theta \cdot \sin \theta \left( {\sin \theta - \cos \theta } \right)}}\]
So, by cutting the equal terms in the numerator and denominator, we obtain
=\[\dfrac{{\left( {{{\sin }^2}\theta + {{\cos }^2}\theta + \sin \theta \cdot \cos \theta } \right)}}{{\cos \theta \cdot \sin \theta }}\]
= \[\dfrac{{1 + \sin \theta \cdot \cos \theta }}{{\cos \theta \cdot \sin \theta }}\] \[\left[ {\because {{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right]\]
Now, by separating the terms, we obtain
= \[\dfrac{1}{{\cos \theta \cdot \sin \theta }} + \dfrac{{\sin \theta \cdot \cos \theta }}{{sin\theta \cdot \cos \theta }}\]
By cutting out the equal terms,
= \[\sec \theta \cdot cosec\theta + 1\] , which can also be written as,
=\[1 + \sec \theta \cdot cosec\theta \]
=RHS
Hence Proved.

(iv) \[\dfrac{{1 + \sec A}}{{\sec A}} = \dfrac{{{{\sin }^2}A}}{{1 - \cos A}}\]
Taking LHS, we get
\[\dfrac{{1 + \sec A}}{{\sec A}}\]
Now, we will make it in terms of cosine.
As we know that, \[\sec A = \dfrac{1}{{\cos A}}\], then,
     \[\]
= \[\dfrac{{1 + \dfrac{1}{{\cos A}}}}{{\dfrac{1}{{\cos A}}}}\]
= \[\dfrac{{\dfrac{{\cos A + 1}}{{\cos A}}}}{{\dfrac{1}{{\cos A}}}}\]
Now, by evaluating it, we obtain
= \[\dfrac{{\cos A + 1}}{{\cos A}} \times \dfrac{{\cos A}}{1}\]
By cutting out equal terms, we get
=\[\cos A + 1\]
Or \[1 + \cos A\]
Now, taking RHS, we get
\[\dfrac{{{{\sin }^2}A}}{{1 - \cos A}}\]
We know that, \[{\sin ^2}A + {\cos ^2}A = 1\]
                        \[ \Rightarrow {\sin ^2}A = 1 - {\cos ^2}A\]
Therefore, it becomes
= \[\dfrac{{1 - {{\cos }^2}A}}{{1 - \operatorname{cosA} }}\]
= \[\dfrac{{{{\left( 1 \right)}^2} - {{\left( {\cos A} \right)}^2}}}{{1 - \cos A}}\]
Now, in the numerator, \[{\left( 1 \right)^2} - {\left( {\cos A} \right)^2}\] represents an identity, i.e., \[{a^2} - {b^2}\]\[ = \left( {a - b} \right)\left( {a + b} \right)\], then it becomes,
= \[\dfrac{{\left( {1 - \cos A} \right)\left( {1 + \cos A} \right)}}{{\left( {1 - \cos A} \right)}}\]
By cutting equal terms, we obtain
=\[1 + \cos A\]
\[\therefore \] LHS=RHS
Hence Proved.

(v) \[\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \cos ecA + \cot A\]
Taking LHS, we get
\[\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}}\]
Now, dividing each term of numerator and denominator by \[\sin A\], we obtain
= \[\dfrac{{\dfrac{{\cos A}}{{\sin A}} - \dfrac{{\sin A}}{{\sin A}} + \dfrac{1}{{\sin A}}}}{{\dfrac{{\cos A}}{{\sin A}} + \dfrac{{\sin A}}{{\sin A}} - \dfrac{1}{{\sin A}}}}\]
Cutting out all the equal terms, we get
= \[\dfrac{{\dfrac{{\cos A}}{{\sin A}} - 1 + \dfrac{1}{{\sin A}}}}{{\dfrac{{\cos A}}{{\sin A}} + 1 - \dfrac{1}{{\sin A}}}}\]
We know that, \[\dfrac{{\cos A}}{{\sin A}} = \cot A\] and \[\dfrac{1}{{\sin A}} = \cos ecA\], then it becomes
= \[\dfrac{{cotA - 1 + cosecA}}{{\cot A + 1 - \cos ecA}}\]
= \[\dfrac{{\left( {\cos ecA + \cot A} \right) - 1}}{{\cot A - \cos ecA + 1}}\]
= \[\dfrac{{\left( {\cos ecA + \cot A} \right) - 1}}{{1 - \cos ecA + \cot A}}\]
As we know that, \[\cos e{c^2}A - {\cot ^2}A = 1\], therefore
= \[\dfrac{{\left( {\cos ecA + \cot A} \right) - \left( {\cos e{c^2}A - {{\cot }^2}A} \right)}}{{1 - \cos ecA + \cot A}}\]
We can write the above equation as
\[\dfrac{{\left( {\cos ecA + \cot A} \right) - \left( {\cos ecA - \cot A} \right)\left( {\cos ecA + \cot A} \right)}}{{1 - \cos ecA + \cot A}}\] by using the identity \[{a^2} - {b^2}\]\[ = \left( {a - b} \right)\left( {a + b} \right)\]
Now, taking \[\cos ecA + \cot A\] common in the numerator, we get
= \[\dfrac{{\left( {\cos ecA + \cot A} \right)\left[ {1 - \left( {\cos ecA - \cot A} \right)} \right]}}{{1 - \cos ecA + \cot A}}\]
= \[\dfrac{{\left( {\cos ecA + \cot A} \right)\left( {1 - \cos ecA + \cot A} \right)}}{{\left( {1 - \cos ecA + \cot A} \right)}}\]
Now, by cutting out the equal terms,
= \[\cos ecA + \cot A\]
=RHS
Hence Proved.

(vi) \[\sqrt {\dfrac{{1 + \sin A}}{{1 - \sin A}}} = \sec A + \operatorname{tanA} \]
Taking LHS, we get
\[\sqrt {\dfrac{{1 + \sin A}}{{1 - \sin A}}} \]
Using rationalization, we obtain
= \[\sqrt {\dfrac{{1 + \sin A}}{{1 - \sin A}} \times \dfrac{{1 + \sin A}}{{1 + \sin A}}} \]
= \[\sqrt {\dfrac{{{{\left( {1 + \sin A} \right)}^2}}}{{\left( {1 - \sin A} \right)\left( {1 + \sin A} \right)}}} \]
= \[\sqrt {\dfrac{{{{\left( {1 + \sin A} \right)}^2}}}{{{{\left( 1 \right)}^2} - {{\left( {\sin A} \right)}^2}}}} \] \[\left[ {\because \left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}} \right]\]

= \[\sqrt {\dfrac{{{{\left( {1 + \sin A} \right)}^2}}}{{1 - {{\sin }^2}A}}} \]
We know that, \[{\sin ^2}A + {\cos ^2}A = 1\]
                       \[ \Rightarrow 1 - {\sin ^2}A = {\cos ^2}A\]
Therefore, it becomes
= \[\sqrt {\dfrac{{{{\left( {1 + \sin A} \right)}^2}}}{{{{\cos }^2}A}}} \]
= \[\sqrt {{{\left( {\dfrac{{1 + \sin A}}{{\cos A}}} \right)}^2}} \]
Now, by cutting out the square and the square root, we obtain
= \[\dfrac{{1 + \sin A}}{{\cos A}}\]
= \[\dfrac{1}{{\cos A}} + \dfrac{{\sin A}}{{\cos A}}\]
= \[\sec A + \tan A\] \[\left[ {\because \dfrac{1}{{\cos A}} = \sec Aand\dfrac{{\sin A}}{{\cos A}} = \tan A} \right]\]

=RHS
Hence Proved.

(vii) \[\dfrac{{\sin \theta - 2{{\sin }^3}\theta }}{{2{{\cos }^3}\theta - \cos \theta }} = \tan \theta \]
Taking LHS, we get
\[\dfrac{{\sin \theta - 2{{\sin }^3}\theta }}{{2{{\cos }^3}\theta - \cos \theta }}\]
Now, taking sine common in the numerator and cosine common in the denominator, we obtain
= \[\dfrac{{\sin \theta \left( {1 - 2{{\sin }^2}\theta } \right)}}{{\cos \theta \left( {2{{\cos }^2}\theta - 1} \right)}}\]
= \[\dfrac{{\sin \theta \left[ {1 - 2\left( {{{\sin }^2}\theta } \right)} \right]}}{{\cos \theta \left( {2{{\cos }^2}\theta - 1} \right)}}\]
=\[\dfrac{{\sin \theta \left[ {1 - 2\left( {1 - {{\cos }^2}\theta } \right)} \right]}}{{\cos \theta \left( {2{{\cos }^2}\theta - 1} \right)}}\] \[\left[ \begin{gathered}
  \because {\sin ^2}\theta + {\cos ^2}\theta = 1 \\
   \Rightarrow {\sin ^2}\theta = 1 - {\cos ^2}\theta \\
\end{gathered} \right]\]

= \[\dfrac{{\sin \theta \left( {1 - 2 + 2{{\cos }^2}\theta } \right)}}{{\cos \theta \left( {2{{\cos }^2}\theta - 1} \right)}}\]
= \[\dfrac{{\sin \theta \left( { - 1 + 2{{\cos }^2}\theta } \right)}}{{\cos \theta \left( {2{{\cos }^2}\theta - 1} \right)}}\]
= \[\dfrac{{\sin \theta \left( {2{{\cos }^2}\theta - 1} \right)}}{{\cos \theta \left( {2{{\cos }^2}\theta - 1} \right)}}\]
By cutting out the equal terms, we get
= \[\dfrac{{\sin \theta }}{{\cos \theta }}\]
As we know that, \[\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta \], therefore it becomes
= \[\tan \theta \]
=RHS
Hence Proved.

(viii) \[{\left( {\sin A + \cos ecA} \right)^2} + {\left( {\cos A + \operatorname{secA} } \right)^2} = 7 + {\tan ^2}A + {\cot ^2}A\]
Taking LHS, we get
\[{\left( {\sin A + \cos ecA} \right)^2} + {\left( {\cos A + \sec A} \right)^2}\] \[\]
Here, \[{\left( {\sin A + \cos ecA} \right)^2}and{\left( {\cos A + \sec A} \right)^2}\] represents an identity, i.e., \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\], therefore it becomes
= \[{\sin ^2}A + 2\sin A \cdot \cos ecA + \cos e{c^2}A + {\cos ^2}A + 2\cos A \cdot \sec A + {\sec ^2}A\]
Since, \[\cos ecA = \dfrac{1}{{\sin A}},\sec A = \dfrac{1}{{\cos A}},\cos e{c^2}A = 1 + {\cot ^2}Aand{\sec ^2}A = 1 + {\tan ^2}A\], therefore it becomes,
= \[{\sin ^2}A + {\cos ^2}A + 2\sin A \cdot \dfrac{1}{{\sin A}} + 1 + {\cot ^2}A + 2\cos A \cdot \dfrac{1}{{\cos A}} + 1 + {\tan ^2}A\]
Now, after cutting out the equal terms and by using the formula \[{\sin ^2}A + {\cos ^2}A = 1\], we get
= \[1 + 2 + 1 + {\cot ^2}A + 2 + 1 + {\tan ^2}A\]
= \[7 + {\tan ^2}A + {\cot ^2}A\]
=RHS
Hence Proved.

(ix) \[\left( {cosecA + \sin A} \right)\left( {\sec A - \cos A} \right) = \dfrac{1}{{\tan A + \cot A}}\]
Taking LHS, we get
\[\left( {cosecA - \sin A} \right)\left( {\sec A - \cos A} \right)\]
Now, we will make it in terms of sine and cosine.
As we know that, \[cosec\theta = \dfrac{1}{{\sin \theta }}\] and \[\sec A = \dfrac{1}{{\cos A}}\], therefore it becomes
= \[\left( {\dfrac{1}{{\sin A}} - \sin A} \right)\left( {\dfrac{1}{{\cos A}} - \cos A} \right)\]
By taking LCM, we get
= \[\left( {\dfrac{{1 - {{\sin }^2}A}}{{\sin A}}} \right)\left( {\dfrac{{1 - {{\cos }^2}A}}{{\cos A}}} \right)\]
= \[\dfrac{{{{\cos }^2}A}}{{\sin A}} \times \dfrac{{{{\sin }^2}A}}{{\cos A}}\] \[\left[ \begin{gathered}
  \because {\sin ^2}A + {\cos ^2}A = 1 \\
   \Rightarrow 1 - {\sin ^2}A = {\cos ^2}A \\
  also,{\sin ^2}A + {\cos ^2}A = 1 \\
   \Rightarrow 1 - {\cos ^2}A = {\sin ^2}A \\
\end{gathered} \right]\]

By cutting out the equal terms, we obtain,
= \[\cos A \cdot \sin A\]

Now, taking RHS, we get
\[\dfrac{1}{{\tan A + \cot A}}\]
We know that, \[\cot A = \dfrac{1}{{\tan A}}\], therefore we obtain
= \[\dfrac{1}{{\tan A + \dfrac{1}{{\tan }}}}\]
By taking LCM in the denominator, we get
= \[\dfrac{1}{{\dfrac{{{{\tan }^2}A + 1}}{{\tan A}}}}\]
As we know that, \[{\tan ^2}A + 1 = {\sec ^2}A\], then it becomes
= \[\dfrac{1}{{\dfrac{{{{\sec }^2}A}}{{\tan A}}}}\]
Now, we obtain
= \[\dfrac{{\tan A}}{{{{\sec }^2}A}}\]
= \[\dfrac{{\dfrac{{\sin A}}{{\cos A}}}}{{\dfrac{1}{{{{\cos }^2}A}}}}\] \[\left[ {\because \tan A = \dfrac{{\sin A}}{{\cos A}}and{{\sec }^2}A = \dfrac{1}{{{{\cos }^2}A}}} \right]\]
= \[\dfrac{{\sin A}}{{\cos A}} \times {\cos ^2}A\]
= \[\sin A \cdot \cos A\]
\[\therefore \]LHS=RHS
Hence Proved.

(x) \[\left[ {\dfrac{{1 + {{\tan }^2}A}}{{1 + {{\cot }^2}A}}} \right] = {\left[ {\dfrac{{1 - \tan A}}{{1 - cotA}}} \right]^2} = {\tan ^2}A\]
Taking LHS, we get
\[\left[ {\dfrac{{1 + {{\tan }^2}A}}{{1 + {{\cot }^2}A}}} \right]\]
We will make it in terms of sine and cosine.
As we know that, \[1 + {\tan ^2}A = {\sec ^2}Aand1 + {\cot ^2}A = \cos e{c^2}A\], then it becomes
= \[\dfrac{{{{\sec }^2}A}}{{\cos e{c^2}A}}\]
= \[\dfrac{{\dfrac{1}{{{{\cos }^2}A}}}}{{\dfrac{1}{{{{\sin }^2}A}}}}\] \[\left[ {\because \sec A = \dfrac{1}{{\cos A}}and\cos ecA = \dfrac{1}{{\sin A}}} \right]\]
= \[\dfrac{1}{{{{\cos }^2}A}} \times \dfrac{{{{\sin }^2}A}}{1}\]
= \[\dfrac{{{{\sin }^2}A}}{{{{\cos }^2}A}}\]
= \[{\tan ^2}A\]
Now, taking RHS, we get
\[{\left[ {\dfrac{{1 - \tan A}}{{1 - \cot A}}} \right]^2}\]
We will make it in terms of sine and cosine.
= \[{\left( {\dfrac{{1 - \dfrac{{\sin A}}{{\cos A}}}}{{1 - \dfrac{{\cos A}}{{\sin A}}}}} \right)^2}\] \[\left[ {\because \tan A = \dfrac{{\sin A}}{{\cos A}}and\cot A = \dfrac{{\cos A}}{{\sin A}}} \right]\]
Taking LCM in the numerator and the denominator, we obtain
= \[{\left( {\dfrac{{\dfrac{{\cos A - \sin A}}{{\cos A}}}}{{\dfrac{{\sin A - \cos A}}{{\sin A}}}}} \right)^2}\]
By further solving it, we get
= \[\dfrac{{{{\left( {\cos A - \sin A} \right)}^2}}}{{{{\cos }^2}A}} \times \dfrac{{{{\sin }^2}A}}{{{{\left( {\sin A - \cos A} \right)}^2}}}\]
= \[\dfrac{{{{\left[ { - \left( {\sin A - \cos A} \right)} \right]}^2}}}{{{{\cos }^2}A}} \times \dfrac{{{{\sin }^2}A}}{{{{\left( {\sin A - \cos A} \right)}^2}}}\]
Now, by cutting out the equal terms and as we know the square of negative sign is positive, we obtain
= \[\dfrac{{{{\sin }^2}A}}{{{{\cos }^2}A}}\]
= \[{\tan ^2}A\]
\[\therefore \]LHS=RHS
Hence Proved.

Note: One should be careful while doing such questions because in these questions we have to use different identities which might be confusing sometimes and before doing these questions one must be clear headed in all aspects. These all parts are quite typical, so one must prove these parts as we did in the solution.