Prove the following identities :

$\dfrac{\text{1 + sin A}}{\text{cosec A - cot A}} - \dfrac{\text{1 - sin A}}{\text{cosec A + cot A}}$ = 2(1 + cot A)

Solving L.H.S. of the equation :

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

By formula,

cosec² A - cot² A = 1

⇒ (1 + sin A)(cosec A + cot A) - (1 - sin A)(cosec A - cot A)

⇒ cosec A + cot A + sin A cosec A + sin A cot A - (cosec A - cot A - sin A cosec A + sin A cot A)

⇒ cosec A - cosec A + cot A + cot A + sin A cosec A + sin A cosec A + sin A cot A - sin A cot A

⇒ 2 cot A + 2 sin A cosec A

⇒ 2 cot A + $2 \text{ sin A} \times \dfrac{1}{\text{sin A}}$

⇒ 2 cot A + 2

⇒ 2(cot A + 1).

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

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