Question

${\sin }^{-1}x-{\sin }^{-1}y=\frac{2\pi }{3};{\cos }^{-1}x-{\cos }^{-1}=\frac{\pi }{3}$.

Answer 1

According to equation,

${\sin }^{-1}x-{\sin }^{-1}y=\frac{2\pi }{3}..\left(1\right)$

${\cos }^{-1}x-{\cos }^{-1}=\frac{\pi }{3}...\left(2\right)$

$(\frac{\pi }{2}-{\sin }^{-1}x)-(\frac{\pi }{2}-{\sin }^{-1}y)=\frac{\pi }{3}$

In equation $\left(2\right){\sin }^{-1}x+{\cos }^{-1}x=-\frac{\pi }{2}$

$\Rightarrow \frac{\pi }{2}-{\sin }^{-1}x-\frac{\pi }{2}+{\sin }^{-1}y=\frac{\pi }{3}$

$\Rightarrow -{\sin }^{-1}x+{\sin }^{-1}y=\frac{\pi }{3}...\left(3\right)$

From equation $\left(1\right)$ and $\left(3\right)$

$2{\sin }^{-1}y=\frac{2\pi }{3}+\frac{\pi }{3}=\pi$

$\Rightarrow {\sin }^{-1}y=\frac{\pi }{2}$

$\Rightarrow y=\sin \frac{\pi }{2}=1..\left(4\right)$

From eqs. $\left(1\right)$ and $\left(4\right)$

${\sin }^{-1}x+{\sin }^{-1}1=\left(\sin \frac{\pi }{2}\right)$

$\Rightarrow {\sin }^{-1}x+\frac{\pi }{2}=\frac{2\pi }{3}$