If y(x)=cot−1(1+sinx+1−sinx1+sinx−1−sinx),x∈(π2,π) ,then dydx at x=5π6 is:

y(x)=cot−1[cosx2+sinx2+sinx2−cosx2cosx2+sinx2−sinx2+cosx2]y(x)=cot−1[tanx2]=π2−x2y'(x)=−12

If y(x)=cot−1(1+sinx+1−sinx1+sinx−1−sinx),x∈(π2,π) ,then dydx at x=5π6 is:

y(x)=cot−1[cosx2+sinx2+sinx2−cosx2cosx2+sinx2−sinx2+cosx2]y(x)=cot−1[tanx2]=π2−x2y'(x)=−12

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If y(x)= $\cot^{- 1} (\frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}), x \in (\frac{π}{2}, π)$ ,then $\frac{d y}{d x}$ at $x = \frac{5 π}{6}$ is:

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Detailed Solution

$y (x) = \cot^{- 1} [\frac{\cos \frac{x}{2} + \sin \frac{x}{2} + \sin \frac{x}{2} - \cos \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2} - \sin \frac{x}{2} + \cos \frac{x}{2}}]$

$y (x) = \cot^{- 1} [\tan \frac{x}{2}] = \frac{π}{2} - \frac{x}{2}$

$y' (x) = \frac{- 1}{2}$

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If y(x)=cot−1(1+sinx+1−sinx1+sinx−1−sinx),x∈(π2,π) ,then dydx at x=5π6 is:

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If y(x)= $\cot^{- 1} (\frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}), x \in (\frac{π}{2}, π)$ ,then $\frac{d y}{d x}$ at $x = \frac{5 π}{6}$ is: