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

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

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

$\dfrac{\text{sin A}}{\text{1 + cos A}} \times \dfrac{1 - \text{cos A}}{1 - \text{cos A }} \\[1em] \dfrac{\text{sin A}(1 - \text{cos A})}{(1 + \text{cos A})(1 - \text{cos A})} \\[1em] \dfrac{\text{sin A}(1 - \text{cos A})}{1 - \text{cos}^2 A} \\[1em] \dfrac{\text{sin A}(1 - \text{cos A})}{\text{sin}^2 A} \\[1em] \dfrac{1 - \text{cos A}}{\text{sin} A}.$

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