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

$\dfrac{\text{cosec A}}{\text{cosec A - 1}} + \dfrac{\text{cosec A}}{\text{cosec A + 1}} = 2 \text{ sec}^2 A$

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

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

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