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

\((1+\sqrt{1+x}) \tan x=1+\sqrt{1-x}\)
\(\Rightarrow \tan x=\frac{1+\sqrt{1-x}}{1+\sqrt{1+x}}\)
Put \(x=\sin \theta\)
\(\therefore \quad \tan x=\frac{1+\sqrt{1-\sin \theta}}{1+\sqrt{1+\sin \theta}}\)
\(\begin{aligned} & =\frac{1+\sqrt{\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)^2}}{1+\sqrt{\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)^2}} \\ & =\frac{1+\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}{1+\cos \frac{\theta}{2}+\sin \frac{\theta}{2}} \\ & =\frac{2 \cos ^2 \frac{\theta}{4}-2 \sin \frac{\theta}{4} \cos \frac{\theta}{4}}{2 \cos ^2 \frac{\theta}{4}+2 \sin \frac{\theta}{4} \cos \frac{\theta}{4}} \\ & =\frac{2 \cos ^2 \frac{\theta}{4}\left(1-\tan \frac{\theta}{4}\right)}{2 \cos ^2 \frac{\theta}{4}\left(1+\tan \frac{\theta}{4}\right)}\end{aligned}\)
\(\therefore \quad \tan x=\tan \left(\frac{\pi}{4}-\frac{\theta}{4}\right) \quad \ldots\left[\because \tan \frac{\pi}{4}=1\right]\)
\(\begin{aligned} & \Rightarrow x=\frac{\pi-\theta}{4} \\ & \Rightarrow \sin 4 x=\sin (\pi-\theta)=\sin \theta=x\end{aligned}\)