Prove the following identity, where the angles involved are acute angles for which the expressions are defined.

$\Big(\dfrac{1 + \text{tan}^2 A}{\text{1 + cot}^2 A}\Big) = \Big(\dfrac{1 - \text{tan A}}{\text{1 - cot A}}\Big)^2$ = tan² A

To prove:

$\Big(\dfrac{1 + \text{tan}^2 A}{\text{1 + cot}^2 A}\Big) = \Big(\dfrac{1 - \text{tan A}}{\text{1 - cot A}}\Big)^2$ = tan² A

Solving L.H.S. of the equation, we get :

$\Rightarrow \Big(\dfrac{1 + \text{tan}^2 A}{\text{1 + cot}^2 A}\Big) \\[1em] \Rightarrow \dfrac{\text{sec}^2 A}{\text{cosec}^2 A} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\text{cos}^2 A}}{\dfrac{1}{\text{sin}^2 A}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A}{\text{cos}^2 A} \\[1em] \Rightarrow \text{tan}^2 A.$

Solving mid-portion, we get :

$\Rightarrow \Big(\dfrac{1 - \text{tan A}}{\text{1 - cot A}}\Big)^2 \\[1em] \Rightarrow \Bigg(\dfrac{1 - \dfrac{\text{sin A}}{\text{cos A}}}{1 - \dfrac{\text{cos A}}{\text{sin A}}}\Bigg)^2 \\[1em] \Rightarrow \Bigg(\dfrac{\dfrac{\text{cos A - sin A}}{\text{cos A}}}{\dfrac{\text{sin A - cos A}}{\text{sin A}}}\Bigg)^2 \\[1em] \Rightarrow \Big(\dfrac{\text{sin A(cos A - sin A)}}{\text{-cos A(cos A - sin A)}}\Big)^2 \\[1em] \Rightarrow (\text{-tan A})^2 \\[1em] \Rightarrow \text{tan}^2 A.$

Hence, proved that $\Big(\dfrac{1 + \text{tan}^2 A}{\text{1 + cot}^2 A}\Big) = \Big(\dfrac{1 - \text{tan A}}{\text{1 - cot A}}\Big)^2$ = tan² A.