SOLUTION : L.H.S. $\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1}\left(\frac{1}{3}\right)=\frac{9}{4}\left[\frac{\pi}{2}-\sin ^{-1}\left(\frac{1}{3}\right)\right]=\frac{9}{4} \cos ^{-1}\left(\frac{1}{3}\right)\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]$
$=\frac{9}{4} \sin ^{-1} \sqrt{1-\left(\frac{1}{3}\right)^2} \quad\left[\because \cos ^{-1} x=\sin ^{-1} \sqrt{1-x^2} \text { for } 0 \leq x \leq 1\right] \\$
$=\frac{9}{4} \sin ^{-1} \sqrt{1-\frac{1}{9}}=\frac{9}{4} \sin ^{-1} \sqrt{\frac{8}{9}}=\frac{9}{4} \sin ^{-1}\left(2 \frac{\sqrt{2}}{3}\right)$
