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

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

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

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

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