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

$\dfrac{1}{1 + \text{sin } θ} + \dfrac{1}{1 - \text{sin } θ}$ = 2 sec² θ

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

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

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

Hence, proved $\dfrac{1}{1 + \text{sin } θ} + \dfrac{1}{1 - \text{sin } θ}$ = 2 sec² θ