inverse trigonometry derivation Prove that : 
$sin^{-1}x+sin^{-1}y = sin^{-1}[x\sqrt{1-y^2}+y\sqrt{1-x^2}]$ 
If -1 $\leq x \leq 1; -1 \leq y \leq 1 $ and $x^2+y^2\leq 1$ or if $xy <0 $ and $x^2+y^2 > 1$
solution : 
Let $sin^{-1}x = \theta$ and $sin^{-1}y = \alpha$ 
Since sin($\theta + \alpha) = sin\theta cos\alpha + cos\theta sin\alpha$
= sin$\theta\sqrt{1-sin^2 \alpha}+\sqrt{1-sin^2\theta}sin\alpha$
=(x$\sqrt{1-y^2}+\sqrt{1-x^2}y)$
=$\theta + \alpha = sin^{1}(x\sqrt{1-y^2}+\sqrt{1-x^2}y)$
=$sin^{-1}x+sin^{-1}y = sin^{-1}[x\sqrt{1-y^2}+y\sqrt{1-x^2}]$
I would like to get an idea on the restriction given in this i.e.
If -1 $\leq x \leq 1; -1 \leq y \leq 1 $ and $x^2+y^2\leq 1$ or if $xy <0 $ and $x^2+y^2 > 1$  ...Request you to please guide on this..
 A: As $x=\sin\theta, \cos^2\theta=1-x^2\implies 1-x^2\ge 0\implies -1\le x\le 1$
Similarly, $-1\le y\le 1$
As the principal value of $\arcsin x$ lies in $\in[-\frac\pi2,\frac\pi2]$
the given relationship will hold iff $-\frac\pi2\le \arcsin x+\arcsin y\le \frac\pi2$
$(1)$ If $x\cdot y<0$ let $x>0,y<0$  
So, $0\le \arcsin x\le \frac\pi2$ and $-\frac\pi2\le \arcsin y\le 0 $ 
$\implies -\frac\pi2\le \arcsin x+\arcsin y\le \frac\pi2$
$(2)$ Else $x\cdot y>0$
$(2a)$ If $x>0,y>0, 0\le \arcsin x, \arcsin y\le \frac\pi2$
$\implies \arcsin x+ \arcsin y\ge 0$
and $\arcsin x+ \arcsin y$ will be $\le \frac\pi2$
if $\arcsin y\le \frac\pi2-\arcsin x $
As $\sin x $ is increasing function in  $\in[0,\frac\pi2]$
$ \sin(\arcsin y)\le \sin(\frac\pi2-\arcsin x) $
$\implies y\le \cos(\arcsin x)=\cos(\arccos\sqrt{1-x^2})=\sqrt{1-x^2}$
$\implies y^2\le 1-x^2\iff x^2+y^2\le 1$
$(2b) x,y<0, -\frac\pi2\le \arcsin x, \arcsin y\le 0$
$\implies \arcsin x+ \arcsin y\le 0$
Put $-x=X>0, -y=Y>0$ in case $(2a)$ to prove $X^2+Y^2\le 1\implies x^2+y^2\le 1 $
