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?

In the other way round, let 
$$\arctan\left(\frac{\sqrt{1 + x} - \sqrt{1-x}}{\sqrt{1 + x} + \sqrt{1-x}} \right) =y$$
$\implies -\dfrac\pi2\le y\le\dfrac\pi2\  \ \ \ (1)$ and $\tan y=\dfrac{\sqrt{1 + x} - \sqrt{1-x}}{\sqrt{1 + x} + \sqrt{1-x}}$
Applying Componendo and dividendo,  $$\dfrac{\sqrt{1 + x}}{\sqrt{1 - x}}=\dfrac{1+\tan y}{1-\tan y}$$
Squaring we get,  $$\dfrac{1+x}{1-x}=\dfrac{1+\tan^2y+2\tan y}{1+\tan^2y-2\tan y}$$
Again applying Componendo and dividendo, $$x=\dfrac{2\tan y}{1+\tan^2y}=\sin2y$$
Now $-\dfrac1{\sqrt2}\le x\le1\implies-\dfrac\pi4\le2y\le\dfrac\pi2\  \ \ \ (2)$ (using $(1)$)
Finally, $2y=\arcsin x$ if $-\dfrac\pi2\le2y\le\dfrac\pi2$ which is satisfied by $(2)$
$\implies2y=\arcsin x=\dfrac\pi2-\arccos x$
