If $\sin^{-1}x+\sin^{-1}y=\frac{2π}{3},\cos^{-1}x-\cos^{-1}y=-\frac{π}{3}$ then the number of values of (x, y) is:
Answer & explanation
Correct answer: option 4
Adding, we get: $\frac{π}{2}+\sin^{-1}y-\cos^{-1}y=\frac{π}{3}$
$⇒π-2\cos^{-1}y=\frac{π}{3}⇒\cos^{-1}y=\frac{π}{3}⇒y=\frac{1}{2}$
$∴\sin^{-1}x+\sin^{-1}\frac{1}{2}=\frac{2π}{3}⇒\sin^{-1}x=\frac{2π}{3}-\frac{π}{6}=\frac{π}{2}$
⇒ x = 1 ⇒ only one solution.