If y(x)=\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right), x \in\left(\frac{\pi}{2}, \pi\right), \text { then } \frac{\mathrm{d} y}{\mathrm{~d} x} \text { at } x=\frac{5 \pi}{6} is :
Option: 1 -\frac{1}{2}
Option: 2 -1
Option: 3 0
Option: 4 \frac{1}{2}

Answers (1)

x \in \left ( \frac{\pi}{4},\pi \right )\\

\Rightarrow \frac{x}{2}\in\left ( \frac{\pi}{4},\frac{\pi}{2} \right )\\

y=\left ( \frac{\sqrt{(\sin\frac{x}{2}+\cos\frac{x}{2})^{2}}+\sqrt{\left ( \sin\frac{x}{2}-\cos\frac{x}{2} \right )^{2}}}{\sqrt{\left ( \sin\frac{x}{2}+\cos\frac{x}{2} \right )^{2}}-\sqrt{\left ( \sin\frac{x}{2}-\cos\frac{x}{2} \right )^{2}}} \right )\\

    = \cot^{-1}\left ( \frac{2\sin\frac{x}{2}}{2\cos\frac{x}{2}} \right )= \cot^{-1}\tan\frac{x}{2}\\

    = \frac{\pi}{2}-\tan^{-1}\tan\frac{x}{2}= \frac{\pi}{2}-\frac{x}{2}\\

\frac{dy}{dx}= \frac{-1}{2}

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