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

$\dfrac{\text{cos A}}{\text{cosec A + 1}} + \dfrac{\text{cos A}}{\text{cosec A - 1}} = \text{2 tan A}$.

Solving L.H.S.,

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

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