Prove

L.H.S. $=\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}$

$=\frac{9}{4}\left(\frac{\pi}{2}-\sin ^{-1} \frac{1}{3}\right)$

$=\frac{9}{4}\left(\cos ^{-1} \frac{1}{3}\right)$....(1)            $\left[\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]$

Now, let $\cos ^{-1} \frac{1}{3}=x$. Then, $\cos x=\frac{1}{3} \Rightarrow \sin x=\sqrt{1-\left(\frac{1}{3}\right)^{2}}=\frac{2 \sqrt{2}}{3}$.

$\therefore x=\sin ^{-1} \frac{2 \sqrt{2}}{3} \Rightarrow \cos ^{-1} \frac{1}{3}=\sin ^{-1} \frac{2 \sqrt{2}}{3}$

$\therefore$ L.H.S. $=\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}=$ R.H.S.