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

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

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

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

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