Prove that :

$\Big(1 + \dfrac{1}{\text{tan}^2 A}\Big)\Big(1 + \dfrac{1}{\text{cot}^2 A}\Big) = \dfrac{1}{\text{sin}^2 A - \text{sin}^4 A}$

To prove:

$\Big(1 + \dfrac{1}{\text{tan}^2 A}\Big)\Big(1 + \dfrac{1}{\text{cot}^2 A}\Big) = \dfrac{1}{\text{sin}^2 A - \text{sin}^4 A}$

Solving L.H.S. of the above equation :

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

Solving R.H.S. of the equation :

$\Rightarrow \dfrac{1}{\text{sin}^2 A - \text{sin}^4 A} \\[1em] \Rightarrow \dfrac{1}{\text{sin}^2 A(1 - \text{sin}^2 A)} \\[1em] \Rightarrow \dfrac{1}{\text{sin}^2 A\text{ cos}^2 A}.$

Since, L.H.S. = R.H.S. = $\dfrac{1}{\text{sin}^2 A\text{ cos}^2 A}.$

Hence, proved that $\Big(1 + \dfrac{1}{\text{tan}^2 A}\Big)\Big(1 + \dfrac{1}{\text{cot}^2 A}\Big) = \dfrac{1}{\text{sin}^2 A - \text{sin}^4 A}$