frac{d}{dx} sqrt{frac{1 + cos 2x}{1 - cos 2x} | vedclass

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Question

(B) Let $y = \sqrt{\frac{1 + \cos 2x}{1 - \cos 2x}}$.Using trigonometric identities $1 + \cos 2x = 2 \cos^2 x$ and $1 - \cos 2x = 2 \sin^2 x$,we get:$

Vedclass · Accepted Answer

$-\csc^2 x$

Vedclass · Answer

(B) Let $y = \sqrt{\frac{1 + \cos 2x}{1 - \cos 2x}}$.Using trigonometric identities $1 + \cos 2x = 2 \cos^2 x$ and $1 - \cos 2x = 2 \sin^2 x$,we get:$y = \sqrt{\frac{2 \cos^2 x}{2 \sin^2 x}} = \sqrt{\cot^2 x} = |\cot x|$.Assuming the expression is defined in the first quadrant,$y = \cot x$.Now,differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(\cot x) = -\csc^2 x$.Thus,the correct option is $B$.

$\frac{d}{dx} \sqrt{\frac{1 + \cos 2x}{1 - \cos 2x}} = $

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