$\tan \left( {{{\sin }^{ - 1}}\left( {\frac{x}{{\sqrt 2 }} + \sqrt {\frac{{1 - {x^2}}}{2}} } \right) - {{\sin }^{ - 1}}x} \right)$=? Let $0 \le x \le \frac{1}{2}$. What is $$\tan \left( {{{\sin }^{ - 1}}\left( {\frac{x}{{\sqrt 2 }} + \sqrt {\frac{{1 - {x^2}}}{2}} } \right) - {{\sin }^{ - 1}}x} \right)$$?
 A: Write $x$ as $\sin\theta$ with $0\le \theta\le \pi/6$. Then your expression equals
$$\tan \left( {{\sin }^{ - 1}}\left( \frac{\sin \theta}{{\sqrt 2 }} + \frac{\cos\theta}{\sqrt 2} \right) - \theta \right) $$
But $\sin \theta + \cos\theta = \sqrt2 \sin(\theta+\pi/4)$ (see here), so the remaining arcsine is just $\sin^{-1}(\sin(\theta+\pi/4))$ and what you're left with is
$$ \tan( \theta+\pi/4 - \theta ) = \tan(\pi/4) = 1 $$
no matter what $x$ is, as long as $\theta+\pi/4 \le \pi/2$.
A: Henning's way is fantastic and ought to be chosen as the correct answer.
Nevertheless, here is another, more naive route.  

We start by noting that, for all $x$ in your interval,
$$\arcsin\left(\frac{\sqrt{1-x^2}+x}{\sqrt 2} \right) = \arcsin (x) + \pi/4$$
To prove this we first rearrange and note that $\arcsin$ is an odd function, meaning that we wish to prove
$$\arcsin\left(\frac{\sqrt{1-x^2}+x}{\sqrt 2} \right)+\arcsin (-x)=\pi/4$$
On the LHS we use the Sum of Arcsines identity to rewrite the LHS as
$$\arcsin\left(\frac{\sqrt{1-x^2}+x}{\sqrt 2} \right)+\arcsin (-x)\\= \arcsin \left(\frac{\sqrt{1-x^2}+x}{\sqrt 2}\sqrt{1-x^2}-x\sqrt{1-\left(\frac{\sqrt{1-x^2}+x}{\sqrt 2}\right)^2}\right)$$
With a little bit of work, we can show that the inside evaluates to $\sqrt{2}/2$,  on the interval $[-\sqrt{2}/2,\sqrt{2}/2]$. We thus get $\tan(\arcsin(\sqrt{2}/2))=\tan(\pi/4)=1$
