To find value of $\sin 4x$ Given $\tan x = \frac { 1+ \sqrt{1+x}}{1+ \sqrt{1-x}}$. i have to find value of $\sin 4x$. i write $\sin 4x=4 \frac{ (1-\tan^2 x)(2 \tan x)}{(1+\tan^2 x)^2}$ but it seems very  complicated to do this? Any other methods?
Thanks 
 A: Write $\tan x=\dfrac{1+\sqrt{1+y}}{1+\sqrt{1-y}}$
As $-1\le y\le1$ for real $\tan x$
WLOG let $y=\cos2u$ where $0\le2u\le\pi$
$$\dfrac{1+\sqrt{1+y}}{1+\sqrt{1-y}}=\dfrac{1+\sqrt2\cos u}{1+\sqrt2\sin u}=\dfrac{\dfrac1{\sqrt2}+\cos u}{\dfrac1{\sqrt2}+\sin u}=\dfrac{2\cos\dfrac{45^\circ +u}2\cos\dfrac{45^\circ-u}2}{2\sin\dfrac{45^\circ +u}2\cos\dfrac{45^\circ-u}2}=\cot\dfrac{45^\circ +u}2$$
So, $\tan x=\tan\left(90^\circ-\dfrac{45^\circ +u}2\right)$
$\implies x=n180^\circ+90^\circ-\dfrac{45^\circ +u}2$ where $n$ is any integer
$\implies\sin4x=-\sin(90^\circ+2u)=-\cos2u$
A: $a=\sqrt{1+y}\qquad b=\sqrt{1-y}\qquad t=\tan(x)=\frac{1+a}{1+b}\qquad \sin(4x)=\frac{4t(1-t^2)}{(1+t^2)^2}$
$a^2+b^2=2$
$a^2-b^2=2y$
Let's have fun:
$\displaystyle{\sin(4x)=4\left(\frac{1+a}{1+b}\right)\frac{(1+b)^2-(1+a)^2}{(1+b)^2}\frac{(1+b)^4}{((1+a)^2+(1+b)^2)^2}}$
$((1+a)^2+(1+b)^2)^2=(1+2a+a^2+1+2b+b^2)^2=4(2+a+b)^2$
$(1+b)^2-(1+a)^2=(2+a+b)(b-a)$
We have a good simplification already : $\displaystyle{\sin(4x)=\frac{(1+a)(1+b)(b-a)}{2+a+b}}$
$(2+a+b)(2-a-b)=(4+2a+2b-2a-a^2-ab-2b-ab-b^2)=2-2ab=2(1-ab)$
$(1-ab)(1+ab)=1-a^2b^2=1-(1-y^2)=y^2$
$2y^2\sin(4x)=(1+a)(1+b)(b-a)(2-a-b)(1+ab)$
$(b-a)(1+ab)=b-a+a(1-y)-b(1+y)=-(a+b)y$
$(1+a)(1+b)(2-a-b)=2+a+b-b^2-a^2-a^2b-ab^2=a+b-b(1+y)-a(1-y)=(a-b)y$
$2y^2\sin(4x)=-(a+b)y(a-b)y=-2y^3$
$$\bbox[10px,border:2px solid]{\sin(4x)=-y}$$
