If y = operatorname{cosec}^{-1}left[frac{sqrt | vedclass

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(A) We know that $\operatorname{cosec}^{-1}(z) = \sin^{-1}(\frac{1}{z})$ for $|z| \geq 1$.Given $y = \operatorname{cosec}^{-1}\left(\frac{\sqrt{x}+1}{

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(A) We know that $\operatorname{cosec}^{-1}(z) = \sin^{-1}(\frac{1}{z})$ for $|z| \geq 1$.Given $y = \operatorname{cosec}^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}ight) + \cos^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}ight)$.Using the identity,we can rewrite the first term as $\sin^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}ight)$.Thus,$y = \sin^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}ight) + \cos^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}ight)$.Using the inverse trigonometric identity $\sin^{-1}(	heta) + \cos^{-1}(	heta) = \frac{\pi}{2}$,we get $y = \frac{\pi}{2}$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{2}) = 0$.

If $y = \operatorname{cosec}^{-1}\left[\frac{\sqrt{x}+1}{\sqrt{x}-1}\right] + \cos^{-1}\left[\frac{\sqrt{x}-1}{\sqrt{x}+1}\right]$,then $\frac{dy}{dx} = $

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