The value of cos≤ft((π)/6 + sin^-1≤ft(-(√3)/2)) is

The correct answer is Option (4) → \(\displaystyle\frac{\sqrt{3}}{2}\)

The principal value branch of \(\sin^{-1}\) is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), so for a negative number the angle is negative:

    \( \displaystyle \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3} \)

Then

    \( \displaystyle \cos\left(\frac{\pi}{6} - \frac{\pi}{3}\right) = \cos\left(-\frac{\pi}{6}\right) \)

    \( \displaystyle = \frac{\sqrt{3}}{2} \)

since cosine is an even function.