Question

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

• A. two
• B. four
• C. zero
• D. none of these

Answer 1

The correct option is D none of these

$y=\cos \left(\frac{\pi }{3}\right)$

$y=\frac{1}{2}$

Hence ${\sin }^{-1}\left(y\right)={\sin }^{-1}\left(\frac{1}{2}\right)$

$=\frac{\pi }{6}$ ...(i)

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

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

${\sin }^{-1}\left(x\right)=\frac{4\pi }{6}-\frac{\pi }{6}$

${\sin }^{-1}\left(x\right)=\frac{3\pi }{6}$

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

$x=1$

$(x,y)$$=(1,\frac{1}{2})$