Question

The value of ${\cos }^{-1}\sqrt{\frac{2}{3}}-{\cos }^{-1}\frac{\sqrt{6}+1}{2\sqrt{3}}$ is equal to

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{6}$

Answer 1

Answer: (D)

$\frac{\pi }{6}$

${\cos }^{-1}\sqrt{\frac{2}{3}}-{\cos }^{-1}\frac{\sqrt{6}+1}{2\sqrt{3}}$

Let ${\cos }^{-1}\sqrt{\frac{2}{3}}=\alpha$ & ${\cos }^{-1}\frac{\sqrt{6}+1}{2\sqrt{3}}=\beta$

Let $\alpha -\beta =m$

$\cos m=\cos (\alpha -\beta )$

$=\cos \alpha \cos \beta +\sin \alpha \sin \beta$

$=\sqrt{\frac{2}{3}}\left(\frac{\sqrt{6}+1}{2\sqrt{3}}\right)+\sqrt{1-\frac{2}{3}}\sqrt{1-\frac{(\sqrt{6}+1{)}^2}{12}}$

$=\frac{\sqrt{2}(\sqrt{6}+1)}{6}+\frac{1}{\sqrt{3}}\frac{\sqrt{12-(\sqrt{6}+1{)}^2}}{2\sqrt{3}}$

$=\frac{\sqrt{2}(\sqrt{6}+1)+\sqrt{12-6-1-2\sqrt{6}}}{6}$

$\Rightarrow \left(\frac{\sqrt{2}(\sqrt{6}+1)+\sqrt{5-2\sqrt{6}}}{6}\right)$

$\Rightarrow \frac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\frac{3\sqrt{3}}{6}$

$m={\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi }{6}$.