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

$\sqrt{\dfrac{\text{sec A - 1}}{\text{sec A + 1}}} + \sqrt{\dfrac{\text{sec A + 1}}{\text{sec A - 1}}} = \text{2 cosec A}$

The L.H.S. of the equation can be written as,

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

Since, L.H.S. = R.H.S. hence, proved that $\sqrt{\dfrac{\text{sec A - 1}}{\text{sec A + 1}}} + \sqrt{\dfrac{\text{sec A + 1}}{\text{sec A - 1}}}$ = 2 cosec A.