If $\sin^{-1}x+\sin^{-1}y=\frac{2π}{3},\cos^{-1}x-\cos^{-1}y=-\frac{π}{3}$ then the number of values of (x, y) is:

None of these

Adding, we get: $\frac{π}{2}+\sin^{-1}y-\cos^{-1}y=\frac{π}{3}$

$⇒π-2\cos^{-1}y=\frac{π}{3}⇒\cos^{-1}y=\frac{π}{3}⇒y=\frac{1}{2}$

$∴\sin^{-1}x+\sin^{-1}\frac{1}{2}=\frac{2π}{3}⇒\sin^{-1}x=\frac{2π}{3}-\frac{π}{6}=\frac{π}{2}$

⇒ x = 1 ⇒ only one solution.