Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $ It is known that
\begin{align}
\arcsin x + \arcsin y  =\begin{cases}
\arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\text{if } x^2+y^2 \le 1 &\text{or} &(x^2+y^2 > 1 &\text{and} &xy< 0);\\
\pi - \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\text{if } x^2+y^2 > 1&\text{and} &0< x,y \le 1;\\
-\pi - \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\text{if } x^2+y^2 > 1&\text{and} &-1\le x,y < 0.
\end{cases}
\end{align}
I tried to prove this myself, have no problem in getting the 'crux' $\arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2})$ part of the RHS, but face trouble in checking the range of that 'crux' under the given conditions.
 A: Let $a=\sin^{-1}x,$ $b=\sin^{-1}y\implies\sin a=x,$ $\sin b=y$ and $a,b\in[-\pi/2,\pi/2]\implies a+b\in[-\pi,\pi]$
$$
\sin(a+b)=\sin a\cos b+\cos a\sin b=x\sqrt{1-y^2}+y\sqrt{1-x^2}\\=\sin\bigg[\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)\bigg]\\
\implies a+b=\color{red}{\sin^{-1}x+\sin^{-1}y=n\pi+(-1)^n\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)}
$$
Case 1: $-\dfrac{\pi}{2}\leq\sin^{-1}x+\sin^{-1}y\leq\dfrac{\pi}{2}$
$$
\color{darkblue}{\sin^{-1}x+\sin^{-1}y=\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)}
$$
$$
\cos(a+b)\geq0\implies \cos a\cos b-\sin a\sin b\geq0\implies \cos a\cos b\geq\sin a\sin b\\
\sqrt{1-x^2}\sqrt{1-y^2}\geq xy\implies\\
a)\ 1-x^2-y^2+x^2y^2-x^2y^2\geq0 \implies x^2+y^2\leq1\\
b)\ 1-x^2-y^2+x^2y^2-x^2y^2<0 \implies x^2+y^2>1, xy<0
$$
Case 2 & 3: $\dfrac{\pi}{2}<\sin^{-1}x+\sin^{-1}y\leq\pi$ and $-\pi\leq\sin^{-1}x+\sin^{-1}y<-\dfrac{\pi}{2}$
$$
\cos(a+b)<0\implies x^2+y^2>1
$$
Case 2-: $\dfrac{\pi}{2}<\sin^{-1}x+\sin^{-1}y\leq\pi$
$$
\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)=\pi-(\sin^{-1}x+\sin^{-1}y)\\
\implies \color{darkblue}{\sin^{-1}x+\sin^{-1}y=\pi-\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)}
$$
$$
\dfrac{\pi}{2}<\sin^{-1}x+\sin^{-1}y\leq\pi \;\&\;-\dfrac{\pi}{2}\leq\sin^{-1}x,\;\sin^{-1}y\leq\dfrac{\pi}{2}\\
\implies 0<\sin^{-1}x,\sin^{-1}y\leq\dfrac{\pi}{2}\implies 0< x,y\leq1
$$
Case 3-: $-\pi\leq\sin^{-1}x+\sin^{-1}y<-\dfrac{\pi}{2}$
$$
\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)=-\pi-(\sin^{-1}x+\sin^{-1}y)\\
\implies \color{darkblue}{\sin^{-1}x+\sin^{-1}y=-\pi-\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)}
$$
$$
-\pi\leq\sin^{-1}x+\sin^{-1}y<-\dfrac{\pi}{2} \;\&\;-\dfrac{\pi}{2}\leq\sin^{-1}x,\;\sin^{-1}y\leq\dfrac{\pi}{2}\\
\implies -\dfrac{\pi}{2}\leq\sin^{-1}x,\sin^{-1}y< 0\implies -1\leq x,y< 0
$$
A: Hope the graph explains the x,y domain ( |x|< 1, |y|< 1 ) and range.

A: Using this, $\displaystyle-\frac\pi2\leq \arcsin z\le\frac\pi2 $ for $-1\le z\le1$
So, $\displaystyle-\pi\le\arcsin x+\arcsin y\le\pi$
Again, $\displaystyle\arcsin x+\arcsin y= 
\begin{cases} 
\\-\pi- \arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2})& \mbox{if } -\pi\le\arcsin x+\arcsin y<-\frac\pi2\\
\arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2}) &\mbox{if } -\frac\pi2\le\arcsin x+\arcsin y\le\frac\pi2 \\ 
\pi- \arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2})& \mbox{if }\frac\pi2<\arcsin x+\arcsin y\le\pi \end{cases} $
and as like other trigonometric ratios are $\ge0$ for the angles in $\left[0,\frac\pi2\right]$
So, $\displaystyle\arcsin z\begin{cases}\text{lies in } \left[0,\frac\pi2\right] &\mbox{if } z\ge0 \\ 
\text{lies in } \left[-\frac\pi2,0\right] & \mbox{if } z<0  \end{cases} $
Case $(i):$
Observe that if $\displaystyle x\cdot y<0\ \ \ \ (1)$ i.e., $x,y$ are of opposite sign, $\displaystyle -\frac\pi2\le\arcsin x+\arcsin y\le\frac\pi2$
Case $(ii):$
If $x>0,y>0$   $\displaystyle \arcsin x+\arcsin y$ will be $\displaystyle \le\frac\pi2$ according as 
$\displaystyle \arcsin x\le\frac\pi2-\arcsin y$
But as $\displaystyle\arcsin y+\arccos y=\frac\pi2,$ we need $\displaystyle  \arcsin x\le\arccos y$
Again as the principal value of inverse cosine ratio lies in $\in[0,\pi],$
$\displaystyle\arccos y=\arcsin(+\sqrt{1-y^2})\implies \arcsin x\le\arcsin\sqrt{1-y^2}$
Now as sine ratio is increasing in $\displaystyle \left[0,\frac\pi2\right],$ we need $\displaystyle x\le\sqrt{1-y^2}\iff x^2\le1-y^2$ as $x,y>0$
$\displaystyle\implies x^2+y^2\le1 \ \ \ \ (2)$
So, $(1),(2)$ are the required condition  for $\displaystyle \arcsin x+\arcsin y\le\frac\pi2$
Case $(iii):$
Now as $\displaystyle-\frac\pi2\arcsin(-u)\le\frac\pi2 \iff -\frac\pi2\arcsin(u)\le\frac\pi2$
$\arcsin(-u)=-\arcsin u$ 
Use this fact to find the similar condition when $x<0,y<0$ setting $x=-X,y=-Y$
A: Take the sine of both sides, and use the angle addition formula, then further simplify it by using the fact that $\cos\arcsin t=\sqrt{\cos^2\arcsin t}=\sqrt{1-\sin^2\arcsin t}=\sqrt{1-t^2}$. Then apply the $\arcsin$ function to both sides, and you're done.
