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If

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

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

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