Prove that $\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1}\left(\frac{1}{3}\right)=\frac{9}{4} \sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)$
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Prove that $\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1}\left(\frac{1}{3}\right)=\frac{9}{4} \sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)$

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

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

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

$=\frac{9}{4} \sin ^{-1} \sqrt{1-\frac{1}{9}}=\frac{9}{4} \sin ^{-1} \sqrt{\frac{9-1}{9}}$

$=\frac{9}{4} \sin ^{-1} \sqrt{\frac{8}{9}}=\frac{9}{4} \sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)=\text { R.H.S. }$

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