Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

$\dfrac{1 - \text{tan}^2 A}{\text{cot}^2 A - 1} = \text{tan}^2 A$

The L.H.S. of the equation can be written as,

$\Rightarrow \dfrac{1 - \dfrac{\text{sin}^2 A}{\text{cos}^2 A}}{\dfrac{\text{cos}^2 A}{\text{sin}^2 A} - 1} \\[1em] \Rightarrow \dfrac{\dfrac{\text{cos}^2 A - \text{sin}^2 A}{\text{cos}^2 A}}{\dfrac{\text{cos}^2 A - \text{sin}^2 A}{\text{sin}^2 A}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A - \text{sin}^2 A}{\text{cos}^2 A - \text{sin}^2 A} \times \dfrac{\text{sin}^2 A}{\text{cos}^2 A} \\[1em] \Rightarrow 1 \times \text{tan}^2 A \\[1em] \Rightarrow \text{tan}^2 A.$

Since, L.H.S. = R.H.S. hence, proved that $\dfrac{1 - \text{tan}^2 A}{\text{cot}^2 A - 1} = \text{tan}^2 A$.