Prove that: 9 8 -9 4 ^ -1 1 3 =9 4 ^ -1 22 3 - Vidyadip

Question

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

✓

Answer

$\text{L.H.S.}=\frac{9\pi}{8}-\frac{9}{4}\sin^{-1}\frac{1}{3}$$=\frac{9}{4}\bigg(\frac{\pi}{2}-\sin^{-1}\frac{1}{3}\bigg)$
$=\frac{9}{4}\bigg(\cos^{-1}\frac{1}{3}\bigg) \dots\dots(1)$ $\bigg[\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\bigg]$
$\text{Now, let}\cos^{-1}\frac{1}{3}=x.\text{Then},\cos x=\frac{1}{3}$ $\Rightarrow\sin x=\sqrt{1-\bigg(\frac{1}{3}\bigg)^2}=\frac{2\sqrt{2}}{3}.$ $\therefore x=\sin^{-1}\frac{2\sqrt{2}}{3}\Rightarrow\cos^{-1}\frac{1}{3}=\sin^{-1}\frac{2\sqrt{2}}{3}$ $\therefore\text{L.H.S.}=\frac{9}{4}\sin^{-1}\frac{2\sqrt{2}}{3}=\text{R.H.S.}$

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