<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] For convenience, write
$$ s := \sin x \qquad c := \cos x \qquad m := 1 +\sqrt{1+x} \qquad n := 1 + \sqrt{1-x}$$
The initial equation can then be rewritten as
$$m s = n c \qquad\to\qquad m^2 s^2 = n^2 c^2 \qquad \to\qquad c^2 = \frac{m^2}{m^2+n^2} \qquad s^2 = \frac{n^2}{m^2 + n^2}$$
Note that $m$ and $n$ are both strictly positive. Since (presumably real) $x$ must lie between $1$ and $-1$, we know that $\cos x \geq 0$; since $\tan x$ must be positive in the original equation, $\sin x$ is, too. Thus, we have ...
$$c = \frac{m}{\sqrt{m^2+n^2}} \qquad s = \frac{n}{\sqrt{m^2+n^2}}$$
Now ...
$$\begin{align}
\sin 4x &= 2 \sin 2x \cos 2x \\[4pt]
&= 4 sc ( 2 c^2 - 1 ) \\[4pt]
&= \frac{4mn}{m^2 + n^2} \left(\frac{2 m^2}{m^2 + n^2} - 1\right) \\[4pt]
&= \frac{4mn(m^2-n^2)}{(m^2 + n^2)^2} \\[4pt]
&= \frac{8 x (3 + 2 \sqrt{1 - x} + 2 \sqrt{1 + x} + \sqrt{1 - x} \sqrt{1 + x})}{8 (3 + 2 \sqrt{1 - x} + 2 \sqrt{1 + x} + \sqrt{1 - x} \sqrt{1 + x})} \\[4pt]
&= x
\end{align}$$ 
</s>