Question

Solve for $x$ and $y$; ${\sin }^{-1}x+{\sin }^{-1}y=\frac{2\pi }{3}$ and ${\cos }^{-1}x-{\cos }^{-1}y=\frac{\pi }{3}$

Answer 1

Adding, $\frac{\pi }{2}+{\sin }^{-1}y-{\cos }^{-1}y=\pi$
$\Rightarrow \pi -2{\cos }^{-1}y=\pi$ or ${\cos }^{-1}y=0\Rightarrow y=1$
$\therefore {\sin }^{-1}x+{\sin }^{-1}1=\frac{2\pi }{3}$
or ${\sin }^{-1}x=\frac{2\pi }{3}-\frac{\pi }{2}=\frac{\pi }{6}\Rightarrow x=\frac{1}{2}$
So, there is one solution.