Prove the following identities :

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

Solving L.H.S. of the equation :

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

By formula,

sin² A = 1 - cos² A

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

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{1}{\text{sin A + cos A}} + \dfrac{1}{\text{sin A - cos A}} = \dfrac{\text{2 sin A}}{\text{1 - 2 cos}^2 A}$.