cos≤ft(π/6-cos^-1(-√3/2)) is equal to:

The correct answer is Option (3) → \(-1/2\)

The principal value of \(\cos^{-1}\) lies in [0, π]. Since \(\cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}\) and \(\frac{5\pi}{6}\) is in [0, π]:

    \( \displaystyle \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \frac{5\pi}{6} \)

    \( \displaystyle \cos\left(\frac{\pi}{6}-\frac{5\pi}{6}\right) = \cos\left(-\frac{2\pi}{3}\right) \)

    \( \displaystyle = \cos\frac{2\pi}{3} = -\frac{1}{2} \)

Tip: \(\cos^{-1}(-x) = \pi-\cos^{-1}x\), so \(\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)\) is \(\frac{5\pi}{6}\), not \(-\frac{\pi}{6}\).