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

Solution :
Adding, $$\frac{\pi }{2} + {\sin ^{ - 1}}y - {\cos ^{ - 1}}y = \frac{\pi }{3}$$
or, $$\pi - 2{\cos ^{ - 1}}y = \frac{\pi }{3}\,\,{\text{or, }}{\cos ^{ - 1}}y = \frac{\pi }{3}$$
$$\eqalign{ & \Rightarrow \,\,y = \frac{1}{2} \cr & \therefore \,\,{\sin ^{ - 1}}x + {\sin ^{ - 1}}\frac{1}{2} = \frac{{2\pi }}{3}\,\,{\text{or,}}\,\,{\sin ^{ - 1}}x = \frac{{2\pi }}{3} - \frac{\pi }{3} = \frac{\pi }{3} \cr & \Rightarrow \,\,x = \frac{{\sqrt 3 }}{2}. \cr} $$
So, there is one solution.