Limits Continuity And Differentiability Question 55
Question: If $ y={{\cot }^{-1}}[ \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} ] $ , where $ 0<x<\frac{\pi }{2} $ , then $ \frac{dy}{dx} $ is equal to
Options:
A) $ \frac{1}{2} $
B) 2
C) $ \sin x+\cos x $
D) $ \sin x-\cos x $
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Answer:
Correct Answer: A
Solution:
$ y={{\cot }^{-1}}[ \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} ] $
$ y={{\cot }^{-1}}[ \begin{aligned} & \sqrt{{{\cos }^{2}}\frac{x}{2}+{{\sin }^{2}}\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}} \\ & \frac{+\sqrt{{{\cos }^{2}}\frac{x}{2}+{{\sin }^{2}}\frac{x}{2}-2\sin \frac{x}{2}\cos \frac{x}{2}}}{ \begin{aligned} & \sqrt{{{\cos }^{2}}\frac{x}{2}+{{\sin }^{2}}\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}} \\ & -\sqrt{{{\cos }^{2}}\frac{x}{2}+{{\sin }^{2}}\frac{x}{2}-2\sin \frac{x}{2}\cos \frac{x}{2}} \\ \end{aligned}} \\ \end{aligned} ] $
$ y={{\cot }^{-1}}[ \frac{\sqrt{{{( cos\frac{x}{2}+\sin \frac{x}{2} )}^{2}}}+\sqrt{{{( \cos \frac{x}{2}-\sin \frac{x}{2} )}^{2}}}}{\sqrt{{{( \cos \frac{x}{2}+\sin \frac{x}{2} )}^{2}}}-\sqrt{{{( \cos \frac{x}{2}-\sin \frac{x}{2} )}^{2}}}} ] $
$ y={{\cot }^{-1}}[ \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}} ] $
$ y={{\cot }^{-1}}[ \frac{2\cos \frac{x}{2}}{2\sin \frac{x}{2}} ]={{\cot }^{-1}}( \cot \frac{x}{2} )=\frac{x}{2} $
$ \frac{dy}{dx}=\frac{1}{2} $
BETA
