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

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

Solving L.H.S.,

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

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