If y = log tan left(frac{x}{2}right) + sin^{- | vedclass

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(D) $y = \log 	an \left(\frac{x}{2}ight) + \sin^{-1}(\cos x)$ Since $\cos x = \sin \left(\frac{\pi}{2} - xight)$,we have: $y = \log 	an \left(\f

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$\operatorname{cosec} x - 1$

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(D) $y = \log 	an \left(\frac{x}{2}ight) + \sin^{-1}(\cos x)$ Since $\cos x = \sin \left(\frac{\pi}{2} - xight)$,we have: $y = \log 	an \left(\frac{x}{2}ight) + \sin^{-1} \left[ \sin \left(\frac{\pi}{2} - xight) ight] = \log 	an \left(\frac{x}{2}ight) + \frac{\pi}{2} - x$ Now,differentiating with respect to $x$: $\frac{dy}{dx} = \frac{1}{	an \left(\frac{x}{2}ight)} \cdot \sec^2 \left(\frac{x}{2}ight) \cdot \frac{1}{2} + 0 - 1$ $\frac{dy}{dx} = \frac{\cos \left(\frac{x}{2}ight)}{\sin \left(\frac{x}{2}ight)} \cdot \frac{1}{\cos^2 \left(\frac{x}{2}ight)} \cdot \frac{1}{2} - 1$ $\frac{dy}{dx} = \frac{1}{2 \sin \left(\frac{x}{2}ight) \cos \left(\frac{x}{2}ight)} - 1$ Using the identity $\sin x = 2 \sin \left(\frac{x}{2}ight) \cos \left(\frac{x}{2}ight)$: $\frac{dy}{dx} = \frac{1}{\sin x} - 1 = \operatorname{cosec} x - 1$

If $y = \log \tan \left(\frac{x}{2}\right) + \sin^{-1}(\cos x)$,then $\frac{dy}{dx} = $

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