Solve this

Question:

If $-\frac{\pi}{2}

Solution:

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

$=\tan ^{-1} \sqrt{\frac{1-\left(1-2 \sin ^{2} x\right)}{1+\left(2 \cos ^{2} x-1\right)}}$

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

$=\tan ^{-1} \sqrt{\tan ^{2} x}$

When $-\frac{\pi}{2}<\mathrm{x}<0, \tan \mathrm{x}$ is negative. So, square root of $\tan ^{2} \mathrm{x}$ in this condition is $-\tan \mathrm{x}$.

So, $y=\tan ^{-1} \sqrt{\tan ^{2} x}$

$=\tan ^{-1}(-\tan x)$

$=-\tan ^{-1}(\tan x)$

$=-x$

And so $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}(-\mathrm{x})$

$=-1$, for $x \in\left(-\frac{\pi}{2}, 0\right)$ (Ans)

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