If
Question:

If $\pi<x<\frac{3 \pi}{2}$, then write the value of $\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}$.

Solution:

We have,

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

$=\frac{|\sin x|}{|\cos x|}$

$=\frac{|\sin x|}{|\cos x|}$

$=\frac{-\sin x}{-\cos x} \quad\left(\because \pi<x<\frac{3 \pi}{2}\right)$

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

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