Simplify: $\cos^{-1} x + \cos^{-1} \left[ \frac{x}{2} + \frac{\sqrt{3 - 3x^2}}{2} \right]; \frac{1}{2} \leq x \leq 1$

$\frac{\pi}{3}$

The correct answer is Option (2) → $\frac{\pi}{3}$ ##

Let $\cos^{-1} x = a, x = \cos a$

$ \text{Now, } \cos^{-1} x + \cos^{-1} \left[ \frac{x}{2} + \frac{\sqrt{3 - 3x^2}}{2} \right]$

$= a + \cos^{-1} \left[ \frac{\cos a}{2} + \frac{\sqrt{3 - 3\cos^2 a}}{2} \right]$

$= a + \cos^{-1} \left[ \frac{\cos a}{2} + \frac{\sqrt{3} \cdot \sqrt{1 - \cos^2 a}}{2} \right]$

$= a + \cos^{-1} \left[ \frac{\cos a}{2}  + \frac{\sqrt{3}}{2} \sin a \right]$

$= a + \cos^{-1} \left[ \cos \frac{\pi}{3} \cos a + \sin \frac{\pi}{3} \sin a \right]$

$= a + \cos^{-1} \left[ \cos \left( \frac{\pi}{3} - a \right) \right]$

$= \frac{\pi}{3}$