Question: If $\pi
Solution:
We have,
$\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}=\sqrt{\frac{2 \sin ^{2} x}{2 \cos ^{2} x}}$
$=\frac{|\sin x|}{|\cos x|}$
$=\frac{|\sin x|}{|\cos x|}$
$=\frac{-\sin x}{-\cos x} \quad\left(\because \pi
$\therefore \sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}=\tan x$
