If y = sin left(2 tan ^{-1} sqrt{frac{1+x}{1- | vedclass

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(B) Let $x = \cos 	heta$,then $	heta = \cos^{-1} x$. Substituting $x = \cos 	heta$ in the expression: $y = \sin \left(2 	an^{-1} \sqrt{\frac{1+\co

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$\frac{-x}{\sqrt{1-x^2}}$

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(B) Let $x = \cos 	heta$,then $	heta = \cos^{-1} x$. Substituting $x = \cos 	heta$ in the expression: $y = \sin \left(2 	an^{-1} \sqrt{\frac{1+\cos 	heta}{1-\cos 	heta}}ight)$ Using the identities $1+\cos 	heta = 2 \cos^2 \frac{	heta}{2}$ and $1-\cos 	heta = 2 \sin^2 \frac{	heta}{2}$: $y = \sin \left(2 	an^{-1} \sqrt{\frac{2 \cos^2 \frac{	heta}{2}}{2 \sin^2 \frac{	heta}{2}}}ight) = \sin \left(2 	an^{-1} \left(\cot \frac{	heta}{2}ight)ight)$ Since $\cot \frac{	heta}{2} = 	an \left(\frac{\pi}{2} - \frac{	heta}{2}ight)$: $y = \sin \left(2 \left(\frac{\pi}{2} - \frac{	heta}{2}ight)ight) = \sin (\pi - 	heta) = \sin 	heta$ Since $\sin 	heta = \sqrt{1 - \cos^2 	heta} = \sqrt{1 - x^2}$: $y = \sqrt{1 - x^2}$ Differentiating with respect to $x$: $\frac{dy}{dx} = \frac{1}{2 \sqrt{1 - x^2}} 	imes \frac{d}{dx}(1 - x^2) = \frac{1}{2 \sqrt{1 - x^2}} 	imes (-2x) = \frac{-x}{\sqrt{1 - x^2}}$

If $y = \sin \left(2 \tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right)$,then $\frac{d y}{d x}$ is equal to

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