Prove that $\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)+\sin ^{-1}\left(\frac{16}{65}\right)=\frac{\pi}{2} \quad$
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Prove that $\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)+\sin ^{-1}\left(\frac{16}{65}\right)=\frac{\pi}{2} \quad$

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$\text { L.H.S. }  =\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)+\sin ^{-1}\left(\frac{16}{65}\right)$

$=\sin ^{-1}\left[\frac{4}{5} \sqrt{1-\frac{25}{169}}+\frac{5}{13} \sqrt{1-\frac{16}{25}}\right]+\sin ^{-1}\left(\frac{16}{65}\right)$

$=\sin ^{-1}\left[\frac{4}{5} \times \frac{12}{13}+\frac{5}{13} \times \frac{3}{5}\right]+\sin ^{-1}\left(\frac{16}{65}\right)$

$=\sin ^{-1}\left[\frac{48}{65}+\frac{15}{65}\right]+\sin ^{-1}\left(\frac{16}{65}\right)=\sin ^{-1}\left(\frac{63}{65}\right)+\sin ^{-1}\left(\frac{16}{65}\right)$

$=\sin ^{-1}\left(\frac{63}{65}\right)+\cos ^{-1}\left(\frac{63}{65}\right)=\frac{\pi}{2}=\text { R.H.S. }$

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