If y = operatorname{Sin}^{-1}left(frac{sqrt{1 | vedclass

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(A) Given $y = \operatorname{Sin}^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)$.We know that $1+\sin x = (

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$-\frac{1}{2}$

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(A) Given $y = \operatorname{Sin}^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}ight)$.We know that $1+\sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$ and $1-\sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$.Substituting these,we get $y = \operatorname{Sin}^{-1}\left(\frac{|\cos \frac{x}{2} + \sin \frac{x}{2}| + |\cos \frac{x}{2} - \sin \frac{x}{2}|}{|\cos \frac{x}{2} + \sin \frac{x}{2}| - |\cos \frac{x}{2} - \sin \frac{x}{2}|}ight)$.Given $\frac{-3\pi}{2} In this interval,$\cos \frac{x}{2}  |\cos \frac{x}{2}|$.Simplifying the expression inside the inverse sine,we get $y = \operatorname{Sin}^{-1}(\cot(\frac{x}{2}))$.Since $\cot(\frac{x}{2}) = 	an(\frac{\pi}{2} - \frac{x}{2})$,we have $y = \operatorname{Sin}^{-1}(	an(\frac{\pi}{2} - \frac{x}{2}))$. This simplifies to $y = \frac{\pi}{2} - (\frac{\pi}{2} - \frac{x}{2}) = \frac{x}{2}$ or similar depending on the branch.However,differentiating $y = \frac{x}{2} + C$ gives $\frac{dy}{dx} = \frac{1}{2}$ or $-\frac{1}{2}$.Given the constraints,the derivative is $-\frac{1}{2}$.

If $y = \operatorname{Sin}^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)$ and $\frac{-3\pi}{2} < x < \frac{-\pi}{2}$,then $\frac{dy}{dx} = $

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