$\frac{{9\pi }}{8} - \frac{9}{4}{\sin ^{ - 1}}\frac{1}{3} = \frac{9}{4}{\sin ^{ - 1}}\frac{{2\sqrt 2 }}{3}$
$\frac{{9\pi }}{8} - \frac{9}{4}{\sin ^{ - 1}}\frac{1}{3} = \frac{9}{4}{\sin ^{ - 1}}\frac{{2\sqrt 2 }}{3}$
Official Solution
VVidaara Team
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NCERT & Exemplar
$\frac{{9\pi }}{8} - \frac{9}{4}{\sin ^{ - 1}}\frac{1}{3}$
$= \frac{9}{4}\left[ {\frac{\pi }{2} - {{\sin }^{ - 1}}\frac{1}{3}} \right] = \frac{9}{4}{\cos ^{ - 1}}\frac{1}{3} = \frac{9}{4}{\sin ^{ - 1}}\left( {\frac{{2\sqrt 2 }}{3}} \right)$
Hence, L.H.S. = R.H.S.
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