If `f(x)=sin^(-1)((sqrt(3))/2x-1/2sqrt(1-x^2)),-1/2lt=xlt=1,t h e nf(x)` is equal to `sin^(-1)(1/2)-sin^(-1)(x)` `sin^(-1)x-pi/6` `sin^(-1)x+pi/6` (d) none of these

Options

A. `sin^(-1) ((1)/(2)) - sin^(-1) (x)`
B. `sin^(-1) x - (pi)/(6)`
C. `sin^(-1) x + (pi)/(6)`
D. none of these

Answer

B

Explanation

Let `x = sin theta, " where " -(1)/(2) le x le 1`
or `-(pi)/(6) le theta le (pi)/(2)`. Then,
`f(x) = sin^(-1) ((sqrt3)/(2) x -(1)/(2) sqrt(1 -x^(2)))`
`= sin^(-1) ((sqrt3)/(2) sin theta -(1)/(2) cos theta)`
`= sin^(-1) (sin (theta - (pi)/(6)))`
`= theta - (pi)/(6) = sin^(-1) x -(pi)/(6) " " [ :' theta - (pi)/(6) in ((-pi)/(3), (pi)/(3))]`