<s>[INST] $\tan \left( {{{\sin }^{ - 1}}\left( {\frac{x}{{\sqrt 2 }} + \sqrt {\frac{{1 - {x^2}}}{2}} } \right) - {{\sin }^{ - 1}}x} \right)$=? Let $0 \le x \le \frac{1}{2}$. What is $$\tan \left( {{{\sin }^{ - 1}}\left( {\frac{x}{{\sqrt 2 }} + \sqrt {\frac{{1 - {x^2}}}{2}} } \right) - {{\sin }^{ - 1}}x} \right)$$?
 [/INST] 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 $\tan(\arcsin(\sqrt{2}/2))=\tan(\pi/4)=1$
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