If π/2 less than x less than π, then √(1 - cos 2x)/(1 + cos 2x) = .............................. .

If π/2 < x < π, Then Find √[(1 − cos 2x)/(1 + cos 2x)]

Question:

\[ \frac{\pi}{2}Solution

Using the identities

\[ 1-\cos 2x=2\sin^2x \] and \[ 1+\cos 2x=2\cos^2x \]

Therefore,

\[ \sqrt{\frac{1-\cos 2x}{1+\cos 2x}} = \sqrt{\frac{2\sin^2x}{2\cos^2x}} \] \[ = \sqrt{\tan^2x} \] \[ = |\tan x| \]

Since

\[ \frac{\pi}{2}\(x\) lies in the second quadrant, where

\[ \tan x<0 \]

Hence,

\[ |\tan x|=-\tan x \]

Therefore,

\[ \sqrt{\frac{1-\cos 2x}{1+\cos 2x}} = -\tan x \]

\[ \boxed{-\tan x} \]

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