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)

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