Question

If $\frac{1}{2} \leq x \leq 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. $none\, of\, these$

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} $
$\begin{bmatrix}\tan\alpha = \frac{r \sin\alpha}{r \cos \alpha}\\ \Rightarrow \alpha = \frac{\pi}{3}\end{bmatrix}$