Prove the following identities :

$\text{cot}^2 A \Big(\dfrac{\text{sec A - 1}}{\text{1 + sin A}}\Big) + \text{sec}^2 A \Big(\dfrac{\text{sin A - 1}}{\text{1 + sec A}}\Big)$ = 0

Solving L.H.S. of above equation :

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

By formula,

sec² A - 1 = tan² A

$\Rightarrow \dfrac{\text{cot}^2 A \text{ tan}^2 A}{\text{(1 + sin A)(sec A + 1)}} + \text{sec}^2 A \Big(\dfrac{\text{sin A - 1}}{\text{1 + sec A}}\Big) \\[1em] \Rightarrow \dfrac{\text{cot}^2 A \times \dfrac{1}{\text{ cot}^2 A}}{\text{(1 + sin A)(sec A + 1)}} + \text{sec}^2 A \Big(\dfrac{\text{sin A - 1}}{\text{1 + sec A}}\Big) \\[1em] \Rightarrow \dfrac{1}{\text{(1 + sin A)(sec A + 1)}} + \text{sec}^2 A \Big(\dfrac{\text{sin A - 1}}{\text{1 + sec A}}\Big) \\[1em] \Rightarrow \dfrac{1 + \text{sec}^2 A\text{(sin A - 1)(1 + sin A)}}{\text{(1 + sin A)(sec A + 1)}} \\[1em] \Rightarrow \dfrac{1 - \text{sec}^2 A\text{(1 - sin A)(1 + sin A)}}{\text{(1 + sin A)(sec A + 1)}} \\[1em] \Rightarrow \dfrac{1 - \text{sec}^2 A\text{(1 - sin}^2 A)}{\text{(1 + sin A)(sec A + 1)}} \\[1em] \Rightarrow \dfrac{1 - \text{sec}^2 A \times \text{cos}^2 A}{\text{(1 + sin A)(sec A + 1)}} \\[1em] \Rightarrow \dfrac{1 - \dfrac{1}{\text{cos}^2 A} \times \text{cos}^2 A}{\text{(1 + sin A)(sec A + 1)}} \\[1em] \Rightarrow \dfrac{1 - 1}{\text{(1 + sin A)(sec A + 1)}} \\[1em] \Rightarrow 0.$

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

Hence, proved that $\text{cot}^2 A \Big(\dfrac{\text{sec A - 1}}{\text{1 + sin A}}\Big) + \text{sec}^2 A \Big(\dfrac{\text{sin A - 1}}{\text{1 + sec A}}\Big)$ = 0.