# Prove that –

\[\left( {\dfrac{{1 + {{\tan }^2}A}}{{1 + {{\cot }^2}A}}} \right) = {\left( {\dfrac{{1 - \tan A}}{{1 - \cot A}}} \right)^2} = {\tan ^2}A\]

Answer

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Hint: Convert the first and the second part of the question individually to the third part . Convert $\cot \theta $ to $\tan \theta $ and then cancel out the equal terms from the numerator and the denominator .

Complete step-by-step answer:

First proving $\left( {\dfrac{{1 + {{\tan }^2}A}}{{1 + {{\cot }^2}A}}} \right) = {\tan ^2}A$

$ \Rightarrow \dfrac{{1 + {{\tan }^2}A}}{{1 + \dfrac{1}{{{{\tan }^2}A}}}}$ = ${\tan ^2}A$ ( since $\cot \theta = \dfrac{1}{{\tan \theta }}$ )

$ \Rightarrow \dfrac{{1 + {{\tan }^2}A}}{{\dfrac{{{{\tan }^2}A + 1}}{{{{\tan }^2}A}}}} = {\tan ^2}A$

$ \Rightarrow $ ${\tan ^2}A = {\tan ^2}A$ ( cancelling out the common terms )

LHS = RHS

Now proving ${\left( {\dfrac{{1 - \tan A}}{{1 - \cot A}}} \right)^2} = {\tan ^2}A$

$ \Rightarrow {\left( {\dfrac{{1 - \tan A}}{{1 - \dfrac{1}{{\tan A}}}}} \right)^2} = {\tan ^2}A$ ( since $\cot \theta = \dfrac{1}{{\tan \theta }}$ )

$ \Rightarrow {\left( {\dfrac{{1 - \tan A}}{{\dfrac{{\tan A - 1}}{{\tan A}}}}} \right)^2} = {\tan ^2}A$

$ \Rightarrow $ ${\left( {\tan A} \right)^2} = {\tan ^2}A$ ( cancelling out the common terms )

$ \Rightarrow {\tan ^2}A = {\tan ^2}A$ ( Since ${\left( {\tan A} \right)^2} = {\tan ^2}A$ )

LHS = RHS

Note : In these questions it is advised to simplify the LHS or the RHS according to their complexity of trigonometric functions . Sometimes proving LHS = RHS needs simplification on both sides of the equation . Remember to convert related trigonometric functions to get to the final result.

Complete step-by-step answer:

First proving $\left( {\dfrac{{1 + {{\tan }^2}A}}{{1 + {{\cot }^2}A}}} \right) = {\tan ^2}A$

$ \Rightarrow \dfrac{{1 + {{\tan }^2}A}}{{1 + \dfrac{1}{{{{\tan }^2}A}}}}$ = ${\tan ^2}A$ ( since $\cot \theta = \dfrac{1}{{\tan \theta }}$ )

$ \Rightarrow \dfrac{{1 + {{\tan }^2}A}}{{\dfrac{{{{\tan }^2}A + 1}}{{{{\tan }^2}A}}}} = {\tan ^2}A$

$ \Rightarrow $ ${\tan ^2}A = {\tan ^2}A$ ( cancelling out the common terms )

LHS = RHS

Now proving ${\left( {\dfrac{{1 - \tan A}}{{1 - \cot A}}} \right)^2} = {\tan ^2}A$

$ \Rightarrow {\left( {\dfrac{{1 - \tan A}}{{1 - \dfrac{1}{{\tan A}}}}} \right)^2} = {\tan ^2}A$ ( since $\cot \theta = \dfrac{1}{{\tan \theta }}$ )

$ \Rightarrow {\left( {\dfrac{{1 - \tan A}}{{\dfrac{{\tan A - 1}}{{\tan A}}}}} \right)^2} = {\tan ^2}A$

$ \Rightarrow $ ${\left( {\tan A} \right)^2} = {\tan ^2}A$ ( cancelling out the common terms )

$ \Rightarrow {\tan ^2}A = {\tan ^2}A$ ( Since ${\left( {\tan A} \right)^2} = {\tan ^2}A$ )

LHS = RHS

Note : In these questions it is advised to simplify the LHS or the RHS according to their complexity of trigonometric functions . Sometimes proving LHS = RHS needs simplification on both sides of the equation . Remember to convert related trigonometric functions to get to the final result.

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