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

(a) \( \cosec x+\cot x \)
(b) \( \cosec x-\cot x \)
(c) \( -\cosec x+\cot x \)
(d) \( -\cosec x-\cot x \)

Solution:

\[ \sqrt{\frac{1+\cos x}{1-\cos x}} = \sqrt{\frac{(1+\cos x)^2}{1-\cos^2 x}} \] \[ = \sqrt{\frac{(1+\cos x)^2}{\sin^2 x}} \] \[ = \left|\frac{1+\cos x}{\sin x}\right| \] \[ = |\cosec x+\cot x| \] Since, \[ \pi Hence, correct option is (d).