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

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

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

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

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