Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $

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 $x=-X,y=-Y$

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-: $-\pi\leq\sin^{-1}x+\sin^{-1}y<-\dfrac{\pi}{2}$

$$ \sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)=-\pi-(\sin^{-1}x+\sin^{-1}y)\\ \implies \color{darkblue}{\sin^{-1}x+\sin^{-1}y=-\pi-\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)} $$ $$ -\pi\leq\sin^{-1}x+\sin^{-1}y<-\dfrac{\pi}{2} \;\&\;-\dfrac{\pi}{2}\leq\sin^{-1}x,\;\sin^{-1}y\leq\dfrac{\pi}{2}\\ \implies -\dfrac{\pi}{2}\leq\sin^{-1}x,\sin^{-1}y< 0\implies -1\leq x,y< 0 $$

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.