Given that

$(1+\sqrt{(1+x)})\tan x=1+\sqrt{(1-x)}$

If $4x$ is in second quadrant, then $\sin 4x$ is equal to

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