Mathematics - Matrix Question with Solution | TestHub

Let ${\cos }^{-1}x=y$ $\Rightarrow x=\cos y$, so that $\frac{1}{2}\le x\le 1$ or $0\le y\le \frac{\pi }{3}$ and $\frac{x}{2}+\frac{1}{2}\sqrt{3-3x^2}=\frac{1}{2}\cos y+\frac{\sqrt{3}}{2}\sin y$ $=\cos \frac{\pi }{3}\cos y+\sin \frac{\pi }{3}\sin y=\cos \left(\frac{\pi }{3}-y\right)$ $\Rightarrow {\cos }^{-1}\left(\frac{x}{2}+\frac{1}{2}\sqrt{3-3x^2}\right)=\frac{\pi }{3}-y$ $\therefore$ the given expression is equal to $y+\frac{\pi }{3}-y,\text{ i.e., }\frac{\pi }{3}$