If \( \frac{\pi}{2} < x < \frac{3\pi}{2} \), then \( \sqrt{\frac{1-\sin x}{1+\sin x}} \) is equal to

(a) \( \sec x-\tan x \)
(b) \( \sec x+\tan x \)
(c) \( \tan x-\sec x \)
(d) none of these

Solution:

\[ \sqrt{\frac{1-\sin x}{1+\sin x}} = \sqrt{\frac{(1-\sin x)^2}{1-\sin^2 x}} \] \[ = \sqrt{\frac{(1-\sin x)^2}{\cos^2 x}} \] \[ = \left|\frac{1-\sin x}{\cos x}\right| \] \[ = \left|\sec x-\tan x\right| \] Since, \[ \frac{\pi}{2}0 \] Therefore, \[ \sqrt{\frac{1-\sin x}{1+\sin x}} = \sec x-\tan x \]

Hence, correct option is (a).