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

$\dfrac{1 + \text{sec A}}{\text{sec A}} = \dfrac{\text{sin}^2 A}{1 - \text{cos A}}$

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

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

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