How can I Show that, $\arcsin\left(\frac{b}{c}\right)-\arcsin\left(\frac{a}{c}\right)=2\arcsin\left(\frac{b-a}{c\sqrt2}\right)$ Where $c^2=a^2+b^2$ is Pythagoras theorem. Sides a,b and c are of a right angle triangle.
Show that,
$$\arcsin\left(\frac{b}{c}\right)-\arcsin\left(\frac{a}{c}\right)=2\arcsin\left(\frac{b-a}{c\sqrt2}\right)$$ 
How do I go about proving this identity? Can anybody help? I know of the arctan.

Here's a trigonograph (with $a$ and $b$ reversed, just because):

$$\begin{align}
2\;\angle X &\;=\; \stackrel{\frown}{PQ} - \stackrel{\frown}{RS} \\[6pt]
\implies \qquad\qquad\qquad 2\;\gamma &\;=\; \alpha - \beta \\[6pt]
\implies \qquad 2\,\arcsin \frac{a-b}{2\sqrt{c}} &\;=\; \arcsin\frac{a}{c} - \arcsin\frac{b}{c}
\end{align}$$
