Question

The value of $\cos^{-1}x + \cos^{-1} \left(\frac{x}{2} + \frac{1}{2} \sqrt{3-3x^{2}}\right) ; \frac{1}{2} \le x \le 1$ is

• A. $- \frac{\pi}{3}$
• B. $\frac{\pi}{3}$
• C. $\frac{3}{\pi}$
• D. $- \frac{3}{\pi}$

Answer 1

Answer: (B)

$\frac{\pi}{3}$

Answer 2

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