The value of ${\sin }^{-1}\left(\frac{2\sqrt{2}}{3}\right)+{\sin }^{-1}\left(\frac{1}{3}\right)$ is equal to

We need to find value of ${\sin }^{-1}\left(\frac{2\sqrt{2}}{3}\right)+{\sin }^{-1}\left(\frac{1}{3}\right)$

Let ${\sin }^{-1}\left(\frac{2\sqrt{2}}{3}\right)=t$

$$\begin{gathered} \Rightarrow \sin t=\frac{2\sqrt{2}}{3} \\ {\sin }^{2}t+{\cos }^{2}t=1 \\ {\left(\frac{2\sqrt{2}}{3}\right)}^2+{\cos }^{2}t=1 \\ \Rightarrow {\cos }^{2}t=\frac{1}{9} \\ \Rightarrow \cos t=\frac{1}{3} \\ \Rightarrow t={\cos }^{-1}\left(\frac{1}{3}\right) \end{gathered}$$

Substituting $t$

$$\begin{gathered} \Rightarrow {\cos }^{-1}\left(\frac{1}{3}\right)+{\sin }^{-1}\left(\frac{1}{3}\right) \\ =\frac{\pi }{2} \end{gathered}$$

So, option C is correct.