If
Question:

If $\frac{\pi}{2}<x<\pi$, then $\sqrt{\frac{1-\sin x}{1+\sin x}}+\sqrt{\frac{1+\sin x}{1-\sin x}}$ is equal to

(a) 2 sec x

(b) −2 sec x

(c) sec x

(d) −sec x

Solution:

(b) −2 sec x

$\sqrt{\frac{1-\sin x}{1+\sin x}}+\sqrt{\frac{1+\sin x}{1-\sin x}}$

$=\sqrt{\frac{(1-\sin x)(1-\sin x)}{(1+\sin x)(1-\sin x)}}+\sqrt{\frac{(1+\sin x)(1+\sin x)}{(1-\sin x)(1+\sin x)}}$

$=\sqrt{\frac{(1-\sin x)^{2}}{1-\sin ^{2} x}}+\sqrt{\frac{(1+\sin x)^{2}}{1-\sin ^{2} x}}$

$=\sqrt{\frac{(1-\sin x)^{2}}{\cos ^{2} x}}+\sqrt{\frac{(1+\sin x)^{2}}{\cos ^{2} x}}$

$=\frac{(1-\sin x)}{-\cos x}+\frac{(1+\sin x)}{-\cos s x} \quad\left[\frac{\pi}{2}<x<\pi\right.$, so $\cos x$ will be negative. $]$

$=-(\sec x-\tan x)-(\sec x+\tan x)$

$=-2 \sec x$