Prove the following identities :

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

Solving L.H.S. of the equation :

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

By formula,

sin² A + cos² A = 1

cos² A = 1 - sin² A

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

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

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