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

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

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

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

Since, RHS = LHS, hence proved that $\dfrac{\text{sin A}}{\text{1 - cos A}} = \text{cosec A} + \text{cot A}$