Prove that :

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

Solving L.H.S. of the equation :

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

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

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