Question

If $\frac{1}{2}\le x\le 1,$ then ${\cos }^{-1}x+{\cos }^{-1}\left(\frac{x}{2}+\frac{\sqrt{3-3x^2}}{2}\right)=$

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $noneofthese$

Answer 1

Answer: (C)

${\cos }^{-1}x+{\cos }^{-1}\left(\frac{x}{2}+\frac{\sqrt{3-3x^2}}{2}\right)$
Putting $x=\cos \theta$ in 2nd term, we get
${\cos }^{-1}x+{\cos }^{-1}\left(\frac{\cos \theta }{2}+\frac{\sqrt{3-3{\cos }^2\theta }}{2}\right)$
$={\cos }^{-1}x+{\cos }^{-1}\left(\frac{1}{2}\cos \theta +\frac{\sqrt{3}}{2}\sin \theta \right)$
Now putting $\frac{1}{2}=r\cos \alpha ,\frac{\sqrt{3}}{2}=r\sin \alpha$
$={\cos }^{-1}x+{\cos }^{-1}\left[r\cos \alpha \cos \theta +r\sin \alpha \sin \theta \right]$
$={\cos }^{-1}x+{\cos }^{-1}\left[\cos \left(\alpha -\theta \right)\right]\left[r^2=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\Rightarrow r=1\right]$
$={\cos }^{-1}+\alpha -\theta$
$={\cos }^{-1}x+\left[-{\cos }^{-1}x+\frac{\pi }{3}\right]=\frac{\pi }{3}$
$\left[\begin{array}{c}\tan \alpha =\frac{r\sin \alpha }{r\cos \alpha }\\ \Rightarrow \alpha =\frac{\pi }{3}\end{array}\right]$