Prove the following : $\large\frac{9\pi}{8}-\frac{9}{4}sin^{-1}\bigg(\large\frac{1}{3} \bigg) = \frac{9}{4}sin^{-1}\bigg(\large \frac{2\sqrt 2}{3} \bigg)$

Question

This is Q.No. 12 Misc. of chapter 2