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?
 A: 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 $(1)$
A: Given that $$\tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$$ $$=\tan^{-1}\left(\frac{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}-\sqrt{1-x})}{(\sqrt{1+x}+\sqrt{1-x})(\sqrt{1+x}-\sqrt{1-x})}\right)$$ $$=\tan^{-1}\left(\frac{1+x+1-x-2(\sqrt{1+x})(\sqrt{1-x})}{(\sqrt{1+x})^2-(\sqrt{1-x})^2}\right)$$ $$=\tan^{-1}\left(\frac{2-2\sqrt{1-x^2}}{1+x-(1-x)}\right)$$$$=\tan^{-1}\left(\frac{1-\sqrt{1-x^2}}{x}\right)$$ Now, let $\color{blue}{x=\sin\theta}$ ($\implies \color{blue}{\theta=\sin^{-1}x=\frac{\pi}{2}-\cos^{-1}x}$ $\forall \color{red}{-1\leq x\leq 1}$) & plug it in the above expression we get $$=\tan^{-1}\left(\frac{1-\sqrt{1-(\sin\theta)^2}}{\sin\theta}\right)$$ $$=\tan^{-1}\left(\frac{1-\cos\theta}{\sin\theta}\right)$$ $$=\tan^{-1}\left(\frac{1-\left(1-2\sin^2\frac{\theta}{2}\right)}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right)$$ $$=\tan^{-1}\left(\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right)$$ $$=\tan^{-1}\left(\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}\right)=\tan^{-1}\left(\tan\frac{\theta}{2}\right)=\frac{\theta}{2}$$ $$=\frac{1}{2}\sin^{-1}x$$ $$=\frac{1}{2}\left(\frac{\pi}{2}-\cos^{-1}x\right)$$ $$=\color{blue}{\frac{\pi}{4}-\frac{1}{2}\cos^{-1}x}$$
A: 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 $(2)$
$\implies2y=\arcsin x=\dfrac\pi2-\arccos x$
