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

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

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

$\Rightarrow \dfrac{\text{1 + cosec A}}{\text{cosec A}} \\[1em] \Rightarrow \dfrac{1 + \dfrac{1}{\text{sin A}}}{\dfrac{1}{\text{sin A}}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{sin A + 1}}{\text{sin A}}}{\dfrac{1}{\text{sin A}}} \\[1em] \Rightarrow \dfrac{\text{sin A(sin A + 1)}}{\text{sin A}} \\[1em] \Rightarrow \text{sin A + 1}$

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

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

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