Difficult inverse tangent identity 
Prove that:
$$\arctan\left(\frac{\sqrt{1 + x} - \sqrt{1-x}}{\sqrt{1 + x} + \sqrt{1-x}} \right) = \frac{\pi}{4} - \frac{1}{2}\arccos(x), -\frac{1}{\sqrt{2}} \le x \le 1$$

I'd multiply the inside of $\arctan$ by the conjugate of the denominator.
I get:
$$\arctan\left(\frac{1 - 1\sqrt{1 - x^2}}{x} \right)$$
But that is still very difficult.
Any HINTS, no solutions?

Let $\dfrac12\arccos x=y\implies x=\cos2y$
and $-\dfrac1{\sqrt2}\le x\le1\implies-\dfrac1{\sqrt2}\le\cos2y\le1$
Using the definition of Principal values of $\arccos,0\le2y\le\dfrac{3\pi}4\iff0\le y\le\dfrac{3\pi}8 \  \ \ \ (1)$
$\implies\sin y,\cos y\ge0$
$\implies\sqrt{1-x}=\sqrt{1-\cos2y}=+\sqrt2\sin y$
$\implies\sqrt{1+x}=\sqrt{1+\cos2y}=+\sqrt2\cos y$
Now $$\arctan\left(\frac{\sqrt{1 + x} - \sqrt{1-x}}{\sqrt{1 + x} + \sqrt{1-x}} \right)=\cdots =\arctan\left[\tan\left(\dfrac\pi4-y\right)\right] $$
which is $\dfrac\pi4-y$ 
if $-\dfrac\pi2\le\dfrac\pi4-y\le\dfrac\pi2\iff\dfrac{3\pi}4\ge y\ge-\dfrac\pi4$ which is satisfied by $(1)$
