Question

The value of ${\cos }^{-1}\left(\sqrt{\frac{2}{3}}\right)-{\cos }^{-1}\left(\frac{\sqrt{6}+1}{2\sqrt{3}}\right)$ 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 $\frac{\pi }{6}$

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

${\cos }^{-1}A-{\cos }^{-1}B={\cos }^{-1}(A\times B+\sqrt{1-A^2}\times \sqrt{1-B^2})$
$={\cos }^{-1}(\sqrt{\frac{2}{3}}\times \frac{\sqrt{6}+1}{2\sqrt{3}}+\sqrt{1-\frac{2}{3}}\sqrt{1-{\left(\frac{\sqrt{6}+1}{2\sqrt{3}}\right)}^2})$
$={\cos }^{-1}(\frac{\sqrt{6}+1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\left(\frac{5-2\sqrt{6}}{2\sqrt{3}}\right))$
$={\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi }{6}$