Evaluate cos ≤ft[ cos^-1 ≤ft( -(√3)/2 ) + (π)/6 ].

The correct answer is Option (3) → $-1$ ##

We have,

$\cos \left[ \cos^{-1} \left( -\frac{\sqrt{3}}{2} \right) + \frac{\pi}{6} \right] = \cos \left[ \cos^{-1} \left( \cos \frac{5\pi}{6} \right) + \frac{\pi}{6} \right] \quad \left[ ∵\cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2} \right]$

$= \cos \left( \frac{5\pi}{6} + \frac{\pi}{6} \right) \quad [∵\cos^{-1} (\cos x) = x; \ x \in [0, \pi]]$

$= \cos \left( \frac{6\pi}{6} \right)$

$= \cos(\pi) = -1$