A) \[4x\]
B) \[2x\]
C) \[x\]
D) None of these
Correct Answer: B
Solution :
\[\tan y=\frac{1+\sqrt{1-x}}{1+\sqrt{1+x}}\] If and\[x=\cos \theta \], then\[\sqrt{1-x}=\sqrt{2}\sin (\theta /2)\] \[\sqrt{1+x}=\sqrt{2}\cdot \cos (\theta /2)\] \[\Rightarrow \] \[\tan y=\frac{\sqrt{2}\left| \frac{1}{\sqrt{2}}+\sin \frac{\theta }{2} \right|}{\sqrt{2}\left| \frac{1}{\sqrt{2}}+\cos \frac{\theta }{2} \right|}=\frac{\sin \frac{\pi }{4}+\sin \frac{\theta }{2}}{\cos \frac{\pi }{4}+\cos \frac{\theta }{2}}\] \[\Rightarrow \] \[\tan y=\frac{2\sin \left( \frac{\pi }{8}+\frac{\theta }{4} \right)\cdot \cos \left( \frac{\pi }{8}-\frac{\theta }{4} \right)}{2\cos \left( \frac{\pi }{8}+\frac{\theta }{4} \right)\cdot \cos \left( \frac{\pi }{8}-\frac{\theta }{4} \right)}\] \[\Rightarrow \] \[\tan y=\tan \left( \frac{\pi }{8}+\frac{\theta }{4} \right)\] \[\Rightarrow \] \[4y=\frac{\pi }{2}+\theta \] \[\Rightarrow \] \[\sin 4y=\cos \theta =x\]
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