Finding the area enclosed by curve defined by $\arcsin x+\arcsin y=\arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2})$ 
If $\arcsin x+\arcsin y=\arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2})$
Then the area represented by the locus of point $(x,y)$
if it is given that  $|x|,|y|\leq 1$

My Try: Put $x=\sin \alpha$ and $y=\sin \beta $
and $\alpha,\beta \in [-90^\circ,90^\circ]$ and $\alpha+\beta \in [-180^\circ,180^\circ]$
$\alpha+\beta = \arcsin(\sin (\alpha+\beta))$ which is possible
when $\alpha+\beta\in [-90^\circ,90^\circ]$
Could some help me what is the area enclosed by its locus.Thanks
 A: We are given the square $Q:=[-1,1]^2$ in the $(x,y)$-plane and are told to determine the domain $D$ consisting of all points $(x,y)\in Q$ satisfying the equation
$$\arcsin x+\arcsin y=\arcsin\bigl(x\sqrt{1-y^2}+y\sqrt{1-x^2}\bigr)\ .\tag{1}$$
To this end we draw in a second figure the square $\hat Q:=\bigl[-{\pi\over2},{\pi\over2}\bigr]^2$ in the $(\alpha,\beta)$-plane and consider the map
$$\psi:\quad \hat Q\to Q,\qquad(\alpha,\beta)\mapsto\left\{\eqalign{x&=\sin\alpha \cr y&=\sin\beta\cr}\right.$$
which maps the square $\hat Q$  bijectively onto  $Q$. We have
$$\psi^{-1}:\quad Q\to\hat Q,\qquad (x,y)\mapsto\left\{\eqalign{\alpha&=\arcsin x \cr \beta&=\arcsin y\cr}\right.\quad.$$
Furthermore one has
$$\sqrt{1-x^2}=\cos\alpha,\quad \sqrt{1-y^2}=\cos\beta\ .$$
The  equation $(1)$  reads in the variables $(\alpha,\beta)\in\hat Q$ as follows:
$$\alpha+\beta=\arcsin\bigl(\sin\alpha\cos\beta+\sin\beta\cos\alpha\bigr)=\arcsin\bigl(\sin(\alpha+\beta)\bigr)\ ,$$
and this can be rewritten as
$$\alpha+\beta=\left\{\eqalign{\alpha+\beta\qquad&\bigl(|\alpha+\beta|\leq{\pi\over2}\bigr)\cr
\pi-(\alpha+\beta)\quad&\bigl(\alpha+\beta\geq{\pi\over2})\cr
-\pi-(\alpha+\beta)\quad&\bigl(\alpha+\beta\leq-{\pi\over2}\bigr)\ .\cr}\right.$$
When $|\alpha+\beta|\leq{\pi\over2}$ this requires nothing. If $\alpha+\beta\geq{\pi\over2}$ this says that $\alpha+\beta={\pi\over2}$, and if $\alpha+\beta\leq-{\pi\over2}$ this says that $\alpha+\beta=-{\pi\over2}$. 
We therefore obtain the desired domain $\hat D\subset\hat Q$ by cutting off the two triangles from $\hat Q$ on which $|\alpha+\beta|>{\pi\over2}$. In the original $(x,y)$-figure we obtain the desired domain $D\subset Q$ by cutting off the $\psi$-images of these triangles, which are the points in the first and third quadrants outside the circle $x^2+y^2=1$. Therefore one has
$${\rm area}(D)=2+{\pi\over2}\ .$$
