Question

# Prove that â€“$\left( {\dfrac{{1 + {{\tan }^2}A}}{{1 + {{\cot }^2}A}}} \right) = {\left( {\dfrac{{1 - \tan A}}{{1 - \cot A}}} \right)^2} = {\tan ^2}A$

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 .

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.