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

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(B) Given $y = \sin \left(2 	an ^{-1} \sqrt{\frac{1-x}{1+x}}ight)$ and $x = \cos 2 	heta$. Substitute $x = \cos 2 	heta$ into the expression for

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$-\cot 2 	heta$

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(B) Given $y = \sin \left(2 	an ^{-1} \sqrt{\frac{1-x}{1+x}}ight)$ and $x = \cos 2 	heta$. Substitute $x = \cos 2 	heta$ into the expression for $y$: $\frac{1-x}{1+x} = \frac{1-\cos 2 	heta}{1+\cos 2 	heta} = \frac{2 \sin ^2 	heta}{2 \cos ^2 	heta} = 	an ^2 	heta$. Thus,$	an ^{-1} \sqrt{\frac{1-x}{1+x}} = 	an ^{-1}(	an 	heta) = 	heta$. Therefore,$y = \sin (2 	heta)$. Now,differentiate $y$ and $x$ with respect to $	heta$: $\frac{d y}{d 	heta} = \frac{d}{d 	heta} (\sin 2 	heta) = 2 \cos 2 	heta$. $\frac{d x}{d 	heta} = \frac{d}{d 	heta} (\cos 2 	heta) = -2 \sin 2 	heta$. Using the chain rule,$\frac{d y}{d x} = \frac{d y / d 	heta}{d x / d 	heta} = \frac{2 \cos 2 	heta}{-2 \sin 2 	heta} = -\cot 2 	heta$.

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

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