Prove the following identities :

$\sqrt{\dfrac{\text{1 + sin A}}{\text{1 - sin A}}}$ = sec A + tan A

Solving L.H.S. of the equation :

$\Rightarrow \sqrt{\dfrac{\text{1 + sin A}}{\text{1 - sin A}}}$

Multiplying numerator and denominator by $\sqrt{1 + \text{sin A}}$

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

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

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