JEE MAIN - Mathematics (2021 - 27th August Evening Shift - No. 11)
If $$y(x) = {\cot ^{ - 1}}\left( {{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} } \over {\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right),x \in \left( {{\pi \over 2},\pi } \right)$$, then $${{dy} \over {dx}}$$ at $$x = {{5\pi } \over 6}$$ is :
$$ - {1 \over 2}$$
$$-$$1
$${1 \over 2}$$
0
Explanation
We have,
$$ y(x)=\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right) $$
$$ =\cot ^{-1} \frac{\left|\cos \frac{x}{2}+\sin \frac{x}{2}\right|+\left|\cos \frac{x}{2}-\sin \frac{x}{2}\right|}{\left|\cos \frac{x}{2}+\sin \frac{x}{2}\right|-\left|\cos \frac{x}{2}-\sin \frac{x}{2}\right|} $$
$$ \left[\text { as } \cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}=1\right. $$
and $\left.\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2}\right]$
$$ \begin{aligned} & =\cot ^{-1}\left(\frac{\cos \frac{x}{2}+\sin \frac{x}{2}+\sin \frac{x}{2}-\cos \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}-\sin \frac{x}{2}+\cos \frac{x}{2}}\right) \forall x \in\left(\frac{\pi}{2}, \pi\right) \\ & \end{aligned} $$
$$ =\cot ^{-1}\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right)=\cot ^{-1}\left(\tan \frac{x}{2}\right) $$
$$ =\frac{\pi}{2}-\tan ^{-1}\left(\tan \frac{x}{2}\right) $$
$$ \begin{aligned} & \therefore y^{\prime}(x)=\frac{\pi}{2}-\frac{x}{2} \\\\ & \Rightarrow \frac{d y}{d x}=y^{\prime}(x)=-\frac{1}{2}=y^{\prime}\left(\frac{5 \pi}{6}\right) \end{aligned} $$
$$ y(x)=\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right) $$
$$ =\cot ^{-1} \frac{\left|\cos \frac{x}{2}+\sin \frac{x}{2}\right|+\left|\cos \frac{x}{2}-\sin \frac{x}{2}\right|}{\left|\cos \frac{x}{2}+\sin \frac{x}{2}\right|-\left|\cos \frac{x}{2}-\sin \frac{x}{2}\right|} $$
$$ \left[\text { as } \cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}=1\right. $$
and $\left.\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2}\right]$
$$ \begin{aligned} & =\cot ^{-1}\left(\frac{\cos \frac{x}{2}+\sin \frac{x}{2}+\sin \frac{x}{2}-\cos \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}-\sin \frac{x}{2}+\cos \frac{x}{2}}\right) \forall x \in\left(\frac{\pi}{2}, \pi\right) \\ & \end{aligned} $$
$$ =\cot ^{-1}\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right)=\cot ^{-1}\left(\tan \frac{x}{2}\right) $$
$$ =\frac{\pi}{2}-\tan ^{-1}\left(\tan \frac{x}{2}\right) $$
$$ \begin{aligned} & \therefore y^{\prime}(x)=\frac{\pi}{2}-\frac{x}{2} \\\\ & \Rightarrow \frac{d y}{d x}=y^{\prime}(x)=-\frac{1}{2}=y^{\prime}\left(\frac{5 \pi}{6}\right) \end{aligned} $$
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