Prove that :

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

Solving L.H.S. of the equation :

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

By formula,

cos² A = 1 - sin² A

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

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

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