Question

The number of solution(s) of $(x,y)$ so that ${\sin }^{-1}x+{\sin }^{-1}y=\frac{2\pi }{3}$,${\cos }^{-1}x-{\cos }^{-1}y=\frac{\pi }{3}$is

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

The correct option is A $1$

Given : ${\sin }^{-1}x+{\sin }^{-1}y=\frac{2\pi }{3},{\cos }^{-1}x-{\cos }^{-1}y=\frac{\pi }{3}$
Adding both equations
$\frac{\pi }{2}+{\sin }^{-1}y-{\cos }^{-1}y=\frac{\pi }{3}+\frac{2\pi }{3}=\pi \Rightarrow {\sin }^{-1}y-{\cos }^{-1}y=\frac{\pi }{2}\Rightarrow \frac{\pi }{2}-2{\cos }^{-1}y=\frac{\pi }{2}\Rightarrow {\cos }^{-1}y=0\therefore y=1,x=\frac{1}{2}$