Question
Prove that:
$\frac{9\pi}{8}-\frac{9}{4}\sin^{-1}\frac{1}{3}=\frac{9}{4}\sin^{-1}\frac{2\sqrt{2}}{3}$
$\frac{9\pi}{8}-\frac{9}{4}\sin^{-1}\frac{1}{3}=\frac{9}{4}\sin^{-1}\frac{2\sqrt{2}}{3}$
$=\frac{9}{4}\bigg(\frac{\pi}{2}-\sin^{-1}\frac{1}{3}\bigg)$
$=\frac{9}{4}\bigg(\cos^{-1}\frac{1}{3}\bigg) \dots\dots(1)$ $\bigg[\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\bigg]$
$\text{Now, let}\cos^{-1}\frac{1}{3}=x.\text{Then},\cos x=\frac{1}{3}$ $\Rightarrow\sin x=\sqrt{1-\bigg(\frac{1}{3}\bigg)^2}=\frac{2\sqrt{2}}{3}.$ $\therefore x=\sin^{-1}\frac{2\sqrt{2}}{3}\Rightarrow\cos^{-1}\frac{1}{3}=\sin^{-1}\frac{2\sqrt{2}}{3}$ $\therefore\text{L.H.S.}=\frac{9}{4}\sin^{-1}\frac{2\sqrt{2}}{3}=\text{R.H.S.}$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.