Question

If $f\left(x\right)=si{n}^{-1}(\frac{\sqrt{3}}{2}x-\frac{1}{2}\sqrt{1-x^2}),-\frac{1}{2}\le x\le 1$, then $f\left(x\right)$ is equal to

• A. $si{n}^{-1}\left(\frac{1}{2}\right)-si{n}^{-1}\left(x\right)$
• B. $si{n}^{-1}x-\frac{\pi }{6}$
• C. $si{n}^{-1}x+\frac{\pi }{6}$
• D. $noneofthese$

Answer 1

Answer: (B)

$si{n}^{-1}x-\frac{\pi }{6}$

$si{n}^{-1}(x\frac{\sqrt{3}}{2}-\sqrt{1-x^2}\frac{1}{2})$
$=si{n}^{-1}\left(x\right)-si{n}^{-1}\left(\frac{1}{2}\right)$
$=si{n}^{-1}\left(x\right)-\frac{\pi }{6}$