frac{d}{dx} left[ log sqrt{frac{1 - cos x}{1 | vedclass

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(B) Let $y = \log \sqrt{\frac{1 - \cos x}{1 + \cos x}}$.Using the trigonometric identities $1 - \cos x = 2 \sin^2 \frac{x}{2}$ and $1 + \cos x = 2 \co

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$	ext{cosec } x$

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(B) Let $y = \log \sqrt{\frac{1 - \cos x}{1 + \cos x}}$.Using the trigonometric identities $1 - \cos x = 2 \sin^2 \frac{x}{2}$ and $1 + \cos x = 2 \cos^2 \frac{x}{2}$,we get:$y = \log \sqrt{\frac{2 \sin^2 (x/2)}{2 \cos^2 (x/2)}} = \log \sqrt{	an^2 \frac{x}{2}} = \log \left( 	an \frac{x}{2} ight)$.Now,differentiating with respect to $x$ using the chain rule:$\frac{dy}{dx} = \frac{1}{	an (x/2)} \cdot \sec^2 \frac{x}{2} \cdot \frac{1}{2}$.$\frac{dy}{dx} = \frac{\cos (x/2)}{\sin (x/2)} \cdot \frac{1}{\cos^2 (x/2)} \cdot \frac{1}{2} = \frac{1}{2 \sin (x/2) \cos (x/2)}$.Using the identity $\sin x = 2 \sin (x/2) \cos (x/2)$,we have:$\frac{dy}{dx} = \frac{1}{\sin x} = 	ext{cosec } x$.

$\frac{d}{dx} \left[ \log \sqrt{\frac{1 - \cos x}{1 + \cos x}} \right] = $

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