If `asin^(-1)x-bcos^(-1)x=c ,` then `asin^(-1)x+bcos^(-1)x` is equal to 0 (b) `(pia b+c(b-a))/(a+b)` `pi/2` (d) `(pia b+c(a-b))/(a+b)`

Options

A. 0
B. `(pi ab + c (b -a))/(a + b)`
C. `(pi)/(2)`
D. `(pi ab + c (a -b))/(a + b)`

Answer

D

Explanation

`a sin^(-1) x - b cos^(-1) x = c`
We have `b sin^(-1) x + b cos^(-1) x = (bpi)/(2)`
Adding `(a + b) sin^(-1) x = (bpi)/(2) + c`
`rArr sin^(-1) x = (((bpi)/(2)) + c)/(a + b) = (b pi + 2c)/(2(a + b))`
`:. cos^(-1) x = (pi)/(2) - (b pi + 2c)/(2(a + b)) = (pi a - 2c)/(2(a + b))`
`rArr a sin^(-1) x + b cos^(-1) x = (pi ab + c(a - b))/(a + b)`