If `sin^(-1)x+sin^(-1)y=(2pi)/3` and `cos^(-1)x-cos^(-1)y=-pi/3` then the number of values of `(x,y)` is

Answer

`x = (1)/(2), y = 1`

Explanation

Given equation are
`sin^(-1) x + sin^(-1) y = (2pi)/(3)`
`cos^(-1) x - cos^(-1) y = (pi)/(3)`
`rArr ((pi)/(2) - sin^(-1) x) - ((pi)/(2) - sin^(-1) y) = (pi)/(3)`
Let `sin^(-1) x = A`
`sin^(-1) y = B`
Then Eqs. (i) and (ii) become
`A + B = (2pi)/(3)`
`A - B = -(pi)/(3)`
Solving Eqs. (iii) and (iv), we get
`A = (pi)/(6), B = (pi)/(2)`
`rArr sin^(-1) x = (pi)/(6), sin^(-1) y = (pi)/(2)`
`rArr x = (1)/(2) and y = 1`