A) \[\frac{x-\pi }{2}\]
B) \[\frac{\pi -x}{2}\]
C) \[\frac{3\pi -x}{2}\]
D) None of these
Correct Answer: B
Solution :
[b] \[\sqrt{1+\sin x}=\sin \frac{x}{2}+\cos \frac{x}{2}\] \[\sqrt{1-\sin x}=\sin \frac{x}{2}-\cos \frac{x}{2}\] \[\left[ for\frac{\pi }{4}\le \frac{x}{2}\le \frac{\pi }{2}\sin \frac{\pi }{2}\ge \cos \frac{x}{2} \right]\] \[\therefore \]the expression is \[{{\cot }^{-1}}\left( \frac{\sin \frac{x}{2}+\cos \frac{x}{2}+\sin \frac{x}{2}-\cos \frac{x}{2}}{\sin \frac{x}{2}+\cos \frac{x}{2}-\sin \frac{x}{2}+\cos \frac{x}{2}} \right)\] \[={{\cot }^{-1}}\left( \tan \frac{x}{2} \right)={{\cot }^{-1}}\cot \left( \frac{\pi }{2}-\frac{x}{2} \right)=\frac{\pi -x}{2}\]
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