# If $x\in(\pi,2\pi)$ and $\large\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}$$=\cot\big(a+\large\frac{x}{2}\big) then 'a' is equal to (a)\large\frac{\pi}{4}$$\qquad(b)\;\large\frac{\pi}{2}$$\qquad(c)\;\large\frac{\pi}{3}$$\qquad(d)\;None\;of\;these$

$\sqrt{1+\cos x}=\sqrt{2\cos^2\large\frac{x}{2}}=\sqrt 2\mid \cos\large\frac{x}{2}\mid$
$\sqrt{1-\cos x}=\sqrt{2\sin^2\large\frac{x}{2}}=\sqrt 2\mid \sin\large\frac{x}{2}\mid$
$\large\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}=\large\frac{\mid \cos\Large\frac{x}{2}\mid+\mid\sin\Large\frac{x}{2}\mid}{\mid\cos\Large\frac{x}{2}\mid-\mid \sin\Large\frac{x}{2}\mid}$
$\qquad\qquad\qquad\qquad=\large\frac{\cos\Large\frac{x}{2}-\sin\Large\frac{x}{2}}{\cos\Large\frac{x}{2}+\sin\Large\frac{x}{2}}$
$\qquad\qquad\qquad\qquad=\large\frac{1-\tan\Large\frac{x}{2}}{1+\tan\Large\frac{x}{2}}$
$\qquad\qquad\qquad\qquad=\tan\big(\large\frac{\pi}{4}-\frac{x}{2}\big)$
$\qquad\qquad\qquad\qquad=\cot\big(\large\frac{\pi}{2}-\big(\large\frac{\pi}{4}+\frac{x}{2}\big)\big)$
$\qquad\qquad\qquad\qquad=\cot\big(\large\frac{\pi}{4}+\large\frac{x}{2}\big)$
$\Rightarrow a=\large\frac{\pi}{4}$
Hence (a) is the correct option.