Question

If $(1+\sqrt{1+x}\tan y=1+\sqrt{1-x})$, then $\sin 4y$ is equal to

• A. $4x$
• B. $2x$
• C. $x$
• D. None of these

Answer 1

The correct option is C $x$

$\tan y=\frac{1+\sqrt{1-x}}{1+\sqrt{1+x}}$
If an $x=\cos \theta$, then $\sqrt{1-x}=\sqrt{2}\sin \left(\theta /2\right)$
$\sqrt{1+x}=\sqrt{2}\cdot \cos \left(\theta /2\right)$
$\Rightarrow \tan y=\frac{\sqrt{2}|\frac{1}{\sqrt{2}}+\sin \frac{\theta }{2}|}{\sqrt{2}|\frac{1}{\sqrt{2}}+\cos \frac{\theta }{2}|}=\frac{\sin \frac{\pi }{4}+\sin \frac{\theta }{2}}{\cos \frac{\pi }{4}+\cos \frac{\theta }{2}}$
$\Rightarrow \tan y=\frac{2\sin (\frac{\pi }{8}+\frac{\theta }{4})\cdot \cos (\frac{\pi }{8}-\frac{\theta }{4})}{2\cos (\frac{\pi }{8}+\frac{\theta }{4})\cdot \cos (\frac{\pi }{8}-\frac{\theta }{4})}$
$\Rightarrow \tan y=\tan (\frac{\pi }{8}+\frac{\theta }{4})$
$\Rightarrow 4y=\frac{\pi }{2}+\theta$
$\Rightarrow \sin 4y=\cos \theta =x$.