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

(1 - cos A)(1 + sec A) = tan A sin A.

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

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

Since, L.H.S. = R.H.S. hence proved that (1 - cos A)(1 + sec A) = tan A sin A.