Question

If ${\sin }^{-1}x+{\sin }^{-1}y=\frac{2\pi }{3}$ and ${\cos }^{-1}x-{\cos }^{-1}y=\frac{\pi }{3}$, where $x\in (0,\frac{1}{2})$, then the number of value(s) of $y$ satisfying the equation is

Answer 1

${\sin }^{-1}x+{\sin }^{-1}y=\frac{2\pi }{3}...\left(1\right){\cos }^{-1}x-{\cos }^{-1}y=\frac{\pi }{3}...\left(2\right)$
Adding $\left(1\right)$ and $\left(2\right)$,
$\frac{\pi }{2}+{\sin }^{-1}y-{\cos }^{-1}y=\pi \Rightarrow {\sin }^{-1}y-{\cos }^{-1}y=\frac{\pi }{2}\Rightarrow \frac{\pi }{2}-{\cos }^{-1}y-{\cos }^{-1}y=\frac{\pi }{2}\Rightarrow 2{\cos }^{-1}y=0\Rightarrow y=1$

From eqn $\left(1\right)$,
${\sin }^{-1}x+{\sin }^{-1}1=\frac{2\pi }{3}\Rightarrow {\sin }^{-1}x+\frac{\pi }{2}=\frac{2\pi }{3}\Rightarrow {\sin }^{-1}x=\frac{\pi }{6}\Rightarrow x=\frac{1}{2}$
But $x\in (0,\frac{1}{2})$
So, there is no solution.