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

$\dfrac{\text{1 + sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{1 + sin A}} = 2\text{ sec A}$

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

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

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