If& ^ - 1 x + ^ - 1 y = (2π)/(3) ~and~ ^ - 1 x - ^ - 1 y = (π)/(3)...

Given,

$\sin^{- 1}\text{x} + \sin^{- 1}\text{y} = \dfrac{2\pi}{3} . . . ( i )$

and~$\cos^{- 1}\text{x} - \cos^{- 1}\text{y} = \dfrac{π}{3} . . . ( ii )$

We know that~$\sin^{- 1}\text{x} + \cos^{- 1}\text{x} = \dfrac{π}{2}$

Using the above concept, we can write~

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

$⇒ - \sin^{- 1}\text{x} + \sin^{- 1}\text{y} = \dfrac{π}{3} . . . ( iii )$~

On adding equation~ \left(i\right) and \left(iii\right)$, we get<br /><br />$\sin^{- 1} \text{y} = \dfrac{\pi}{2} ⇒ \text{y} = 1$<br /><br />On subtracting equation \left(i\right)$~from equation~ \left(iii\right)$, we get<br /><br />$\sin^{- 1} \text{x} = \dfrac{\pi}{6} ⇒ \text{x} = \dfrac{1}{2}$<br /><br />$∴ \left(\text{x, y}\right) = \left(\dfrac{1}{2} \text{,}1\right)$</p>$