A) \[\frac{1}{2}\]
B) 2
C) \[\sin x+\cos x\]
D) \[\sin x-\cos x\]
Correct Answer: A
Solution :
[a] \[y={{\cot }^{-1}}\left[ \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right]\] \[y={{\cot }^{-1}}\left[ \begin{align} & \sqrt{{{\cos }^{2}}\frac{x}{2}+{{\sin }^{2}}\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}} \\ & \frac{+\sqrt{{{\cos }^{2}}\frac{x}{2}+{{\sin }^{2}}\frac{x}{2}-2\sin \frac{x}{2}\cos \frac{x}{2}}}{\begin{align} & \sqrt{{{\cos }^{2}}\frac{x}{2}+{{\sin }^{2}}\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}} \\ & -\sqrt{{{\cos }^{2}}\frac{x}{2}+{{\sin }^{2}}\frac{x}{2}-2\sin \frac{x}{2}\cos \frac{x}{2}} \\ \end{align}} \\ \end{align} \right]\] \[y={{\cot }^{-1}}\left[ \frac{\sqrt{{{\left( cos\frac{x}{2}+\sin \frac{x}{2} \right)}^{2}}}+\sqrt{{{\left( \cos \frac{x}{2}-\sin \frac{x}{2} \right)}^{2}}}}{\sqrt{{{\left( \cos \frac{x}{2}+\sin \frac{x}{2} \right)}^{2}}}-\sqrt{{{\left( \cos \frac{x}{2}-\sin \frac{x}{2} \right)}^{2}}}} \right]\] \[y={{\cot }^{-1}}\left[ \frac{\cos \frac{x}{2}+\sin \frac{x}{2}+\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}-\cos \frac{x}{2}+\sin \frac{x}{2}} \right]\] \[y={{\cot }^{-1}}\left[ \frac{2\cos \frac{x}{2}}{2\sin \frac{x}{2}} \right]={{\cot }^{-1}}\left( \cot \frac{x}{2} \right)=\frac{x}{2}\] \[\frac{dy}{dx}=\frac{1}{2}\]
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