If $-1\leq x, y \leq 1$ and $x\sqrt{1-y^2} + y\sqrt{1-x^2} = 1$, find $x^2+y^2$ Let $x,\, y\in\mathbb
 R,\ -1\leq x,\, y\leq 1$ such that $x\sqrt{1-y^2} + y\sqrt{1-x^2} = 1.$
Find the sum $S = x^2+y^2.$

Let $x=\sin\alpha$ and $y=\sin\beta$, where $\{\alpha,\beta\}\subset[-\frac{\pi}{2},\frac{\pi}{2}].$
Thus, the condition gives $\sin(\alpha+\beta)=1.$
Can you end it now?
Also, we can make squaring twice.
Indeed, the condition gives
$$x^2(1-y^2)+y^2(1-x^2)+2xy\sqrt{(1-x^2)(1-y^2)}=1$$ or
$$2xy\sqrt{(1-x^2)(1-y^2)}=1-x^2-y^2+2x^2y^2,$$ which after squaring again gives
$$1+x^4+y^4-2x^2-2y^2+2x^2y^2=0$$ or
$$(x^2+y^2)^2-2(x^2+y^2)+1=0$$ or
$$x^2+y^2=1.$$
