If `sin^(-1)x-cos^(-1)x=(pi)/(6),"then"`

Options

A. `x = (pi)/(8) + sqrt((1)/(2) - (pi^(2))/(64))`
B. `y = sqrt((1)/(2) - (pi^(2))/(64)) - (pi)/(12)`
C. `x = (pi)/(12) + sqrt((1)/(2) - (pi^(2))/(64))`
D. `y = sqrt((1)/(2) - (pi^(2))/(64)) - (pi)/(8)`

Answer

A, D

Explanation

For the given equation `0 le x, y le 1`
Also `sin^(-1) x + sin^(-1) y = (pi)/(2)`
`rArr sin^(-1)x = cos^(-1) y = sin^(-1) sqrt(1 -y^(2))`
`rArr x = sqrt(1 -y^(2))`
`rArr x^(2) + y^(2) = 1`...(i)
Again, `sin 2x = cos 2y`
`rArr cos ((pi)/(2) - 2x) = cos 2y`
`rArr (pi)/(2) -2x = 2n pi +- 2y`, where `n in I`
`rArr x +- y = (pi)/(4) " " ( :' x, y, in [-1, 1])`..(ii)
From Eqs. (i) and (ii), we get
`x = (pi)/(8) + sqrt((1)/(2) - (pi^(2))/(64))`
and `y = sqrt((1)/(2) -(pi^(2))/(64)) - (pi)/(8)`