If y(x) = cot^{-1}left(frac{sqrt{1+sin x} + s | vedclass

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(A) Given $y(x) = \cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right)$.Since $1 \pm \sin x = \left(\cos

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(A) Given $y(x) = \cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}ight)$.Since $1 \pm \sin x = \left(\cos \frac{x}{2} \pm \sin \frac{x}{2}ight)^2$,we have $\sqrt{1+\sin x} = |\cos \frac{x}{2} + \sin \frac{x}{2}|$ and $\sqrt{1-\sin x} = |\cos \frac{x}{2} - \sin \frac{x}{2}|$.For $x \in \left(\frac{\pi}{2}, \piight)$,$\frac{x}{2} \in \left(\frac{\pi}{4}, \frac{\pi}{2}ight)$. In this interval,$\cos \frac{x}{2} > 0$,$\sin \frac{x}{2} > 0$,and $\sin \frac{x}{2} > \cos \frac{x}{2}$.Thus,$\sqrt{1+\sin x} = \cos \frac{x}{2} + \sin \frac{x}{2}$ and $\sqrt{1-\sin x} = \sin \frac{x}{2} - \cos \frac{x}{2}$.Substituting these into the expression for $y(x)$:$y(x) = \cot^{-1}\left(\frac{(\cos \frac{x}{2} + \sin \frac{x}{2}) + (\sin \frac{x}{2} - \cos \frac{x}{2})}{(\cos \frac{x}{2} + \sin \frac{x}{2}) - (\sin \frac{x}{2} - \cos \frac{x}{2})}ight) = \cot^{-1}\left(\frac{2\sin \frac{x}{2}}{2\cos \frac{x}{2}}ight) = \cot^{-1}(	an \frac{x}{2})$.Using $\cot^{-1}(	an 	heta) = \frac{\pi}{2} - 	heta$,we get $y(x) = \frac{\pi}{2} - \frac{x}{2}$.Therefore,$\frac{dy}{dx} = -\frac{1}{2}$.

If $y(x) = \cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right)$,where $x \in \left(\frac{\pi}{2}, \pi\right)$,then the value of $\frac{dy}{dx}$ at $x = \frac{5\pi}{6}$ is:

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