Prove the following identity, where the angles involved are acute angles for which the expressions are defined.

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

To prove:

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

Solving L.H.S. of the above equation, we get :

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

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

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