<s>[INST] Given ${[1+ {(1+ x)}^{1/2}]×\tan(x) = \left[1+ {(1- x)}^{1/2}\right]}$, find $\sin 4 x$. If 
$${[1+ {(1+ x)}^{1/2}]×\tan(x) = \left[1+ {(1- x)}^{1/2}\right]}$$
then find the value of $\sin(4x)$.
The options given are:
a) $x$
b) $4x$
c) $2x$
I tried applying many trigo identities but none of them is working and the radicals are posing a big problem...need help...Thanks!!
 [/INST] Let: $$t = \tan(x) = \frac{1+\sqrt{1-x}}{1+\sqrt{1+x}}$$
Using the double-angle sine and half-angle tangent formulas:
$$
\sin(4x) \;=\; 2\;\sin(2x)\;\cos(2x) \;=\; 2 \cdot \frac{2 t}{1 + t^2} \cdot \frac{1-t^2}{1+t^2} \;=\; \frac{4t(1-t^2)}{(1+t^2)^2}
$$
Substituting back $t$ in terms of $x$ and simplifying:
$$
\require{cancel}
\begin{align}
\sin(4x) & =\; \frac{2 x (\sqrt{1-x^2}+2 \sqrt{1-x}+2 \sqrt{1+x}+3)}{(\sqrt{1-x}+\sqrt{1+x}+2)^2} \\
 & =\; \frac{2 x (\sqrt{1-x^2}+2 \sqrt{1-x}+2 \sqrt{1+x}+3)}{(1-x) + (1+x) + 4 + 2 \sqrt{(1-x)(1+x)} + 4 \sqrt{1-x} + 4 \sqrt{1+x} } \\
 & =\; \frac{\cancel{2} x \bcancel{(\sqrt{1-x^2}+2 \sqrt{1-x}+2 \sqrt{1+x}+3)}}{\cancel{2}\bcancel{(3 + \sqrt{1-x^2} + \sqrt{1-x}+\sqrt{1+x})} } \\
 & =\; x
\end{align}
$$
</s>