Prove that:
$\sqrt{\frac{\sec\theta + 1}{\sec\theta - 1}} + \sqrt{\frac{\sec\theta - 1}{\sec\theta + 1}} = 2,\text{cosec},\theta$
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Solving the L.H.S.,
$$\begin{aligned}\Rightarrow \sqrt{\frac{\sec θ + 1}{\sec θ - 1}} + \sqrt{\frac{\sec θ - 1}{\sec θ + 1}} \\[1em] \Rightarrow \frac{\sqrt{\sec θ + 1}}{\sqrt{\sec θ - 1}} + \frac{\sqrt{\sec θ - 1}}{\sqrt{\sec θ + 1}} \\[1em] \Rightarrow \frac{\sqrt{\sec θ + 1} \times \sqrt{\sec θ + 1} + \sqrt{\sec θ - 1} \times \sqrt{\sec θ - 1}}{\sqrt{\sec θ - 1} \times \sqrt{\sec θ + 1}} \\[1em] \Rightarrow \frac{(\sec θ + 1) + (\sec θ - 1)}{\sqrt{(\sec θ + 1)(\sec θ - 1)}} \\[1em] \Rightarrow \frac{\sec θ + 1 + \sec θ - 1}{\sqrt{\sec^2 θ - 1}} \\[1em] \Rightarrow \frac{2\sec θ}{\sqrt{\sec^2 θ - 1}} \\[1em] \Rightarrow \frac{2\sec θ}{\sqrt{\tan^2 θ}} \\[1em] \Rightarrow \frac{2\sec θ}{\tan θ} \\[1em] \Rightarrow \frac{\frac{2}{\cos θ}}{\frac{\sin θ}{\cos θ}} \\[1em] \Rightarrow 2 \times \frac{1}{\cos θ} \times \frac{\cos θ}{\sin θ} \\[1em] \Rightarrow \frac{2}{\sin θ} \\[1em] \Rightarrow \text{2 cosec θ}.\end{aligned}$$
Since, L.H.S. = R.H.S.
Hence, proved that $\bm{\sqrt{\frac{\sec θ + 1}{\sec θ - 1}} + \sqrt{\frac{\sec θ - 1}{\sec θ + 1}} = 2,\text{cosec},θ}$.