If

Given for $\frac{\pi}{2}

$\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}$

$\left[\right.$ using identities; $\left.1-\cos 2 x=2 \sin ^{2} x 1+\cos 2 x=2 \cos ^{2} x\right]$

$=\sqrt{\frac{2 \sin ^{2} x}{2 \cos ^{2} x}}$

$=\sqrt{\tan ^{2} x}$

$=|\tan x|$

Since $\frac{\pi}{2}

i. e. $|\tan x|=-\tan x$

$\therefore \sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}=-\tan x$