Prove that $\frac{9 \pi}{8} - \frac{9}{4} \sin^{-1} \frac{1}{3} = \frac{9}{4} \sin^{-1} \frac{2 \sqrt{2}}{3}$.

(A) Consider the $L.H.S. = \frac{9 \pi}{8} - \frac{9}{4} \sin^{-1} \frac{1}{3}$.
Factor out $\frac{9}{4}$ from the expression:
$L.H.S. = \frac{9}{4} \left( \frac{\pi}{2} - \sin^{-1} \frac{1}{3} \right)$.
Using the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,we have $\frac{\pi}{2} - \sin^{-1} \frac{1}{3} = \cos^{-1} \frac{1}{3}$.
So,$L.H.S. = \frac{9}{4} \cos^{-1} \frac{1}{3}$.
Let $\cos^{-1} \frac{1}{3} = \theta$. Then $\cos \theta = \frac{1}{3}$.
Since $\sin^2 \theta = 1 - \cos^2 \theta$,we have $\sin^2 \theta = 1 - (\frac{1}{3})^2 = 1 - \frac{1}{9} = \frac{8}{9}$.
Thus,$\sin \theta = \sqrt{\frac{8}{9}} = \frac{2 \sqrt{2}}{3}$.
Therefore,$\theta = \sin^{-1} \frac{2 \sqrt{2}}{3}$.
Substituting this back into the expression,we get $L.H.S. = \frac{9}{4} \sin^{-1} \frac{2 \sqrt{2}}{3} = R.H.S.$