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

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

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

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

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