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

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 $${\rm area}(D)=2+{\pi\over2}\ .$$