Prove the following identities :

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

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\Big(1 + \dfrac{\text{sin}^2 A}{\text{cos}^2 A}\Big) \times \dfrac{\text{cos A}}{\text{sin A}}}{\dfrac{1}{\text{sin}^2 A}} \\[1em] \Rightarrow \Big(\dfrac{\text{cos}^2 A + \text{sin}^2 A}{\text{cos}^2 A}\Big) \times \dfrac{\text{cos A}}{\text{sin A}} \times \text{sin}^2 A$

By formula,

cos² A + sin² A = 1

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

Since, L.H.S. = R.H.S.

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