If `sin^(-1) : [-1, 1] rarr [(pi)/(2), (3pi)/(2)] and cos^(-1) : [-1, 1] rarr [0, pi]` be two bijective functions, respectively inverse of bijective functions `sin : [(pi)/(2), (3pi)/(2)] rarr [-1, 10 and cos : [0, pi] rarr [-1, 1] " then " sin^(-1) x + cos^(-1) x` is

Options

A. `(pi)/(2)`
B. `pi`
C. `(3pi)/(2)`
D. not a constant

Answer

D

Explanation

Let `sin^(-1) x = theta`
`rArr x = sin theta, (pi)/(2) le theta le (3pi)/(2)`
Now, `cos^(-1) x = cos^(-1) (sin theta)`
`= cos^(-1) (-cos ((3pi)/(2) - theta))`
`= pi - cos^(-1) (cos ((3pi)/(2) - theta))`
`= pi -((3pi)/(2) - theta), " as " 0 le (3pi)/(2) - theta le pi`
`= theta -(pi)/(2) = sin^(-1) x - (pi)/(2)`
Hence, `sin^(-1) x + cos^(-1) x = 2 sin^(-1) x - (pi)/(2)`