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

• Correct answer
  A

  $\frac{1}{\sqrt{3}}$
• B

  $\frac{1}{\sqrt{2}}$
• C

  $\frac{2}{\sqrt{3}}$
• D

  $\frac{1}{2 \sqrt{2}}$

Solution

We have, $\tan ^{-1}\left(\frac{1}{2 \sqrt{2}}\right)+\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\cos ^{-1} x$. Let $\theta_1=\tan ^{-1} \frac{1}{2 \sqrt{2}}$ and $\theta_2=\sin ^{-1} \frac{1}{\sqrt{3}}$. Then, $\tan \theta_1=\frac{1}{2 \sqrt{2}}, \sin \theta_2=\frac{1}{\sqrt{3}}$ and $\begin{gathered} \cos \left(\theta_1+\theta_2\right)=x \\ \Rightarrow x=\cos \theta_1 \cos \theta_2-\sin \theta_1 \sin \theta_2 \\ =\frac{2 \sqrt{2}}{3} \cdot \frac{\sqrt{2}}{\sqrt{3}}-\frac{1}{3} \cdot \frac{1}{\sqrt{3}} \end{gathered}$ $=\frac{4}{3 \sqrt{3}}-\frac{1}{3 \sqrt{3}}=\frac{3}{3 \sqrt{3}}=\frac{1}{\sqrt{3}}$

Hence, the correct answer is option: 'A'.