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

The correct option is D $\frac{\pi }{6}$

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