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

$\dfrac{\text{tan A + sin A}}{\text{tan A - sin A}} = \dfrac{\text{sec A + 1}}{\text{sec A - 1}}.$

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

$\Rightarrow \dfrac{\dfrac{\text{sin A}}{\text{cos A}} + \text{sin A}}{\dfrac{\text{sin A}}{\text{cos A}} - \text{sin A}} \\[1em] = \dfrac{\text{sin A}\Big(\dfrac{1}{\text{cos A}} + 1\Big)}{\text{sin A}\Big(\dfrac{1}{\text{cos A}} - 1\Big)} \\[1em] = \dfrac{\Big(\dfrac{1}{\text{cos A}} + 1\Big)}{\Big(\dfrac{1}{\text{cos A}} - 1\Big)} \\[1em] = \dfrac{\text{sec A + 1}}{\text{sec A - 1}}$

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