Question

Prove that $\sin ^{-1}\left(\frac{3}{5}\right)-\cos ^{-1}\left(\frac{12}{13}\right)=\sin ^{-1}\left(\frac{16}{65}\right)$

Answer 1

L.H.S. $=\sin ^{-1}\left(\frac{3}{5}\right)-\cos ^{-1}\left(\frac{12}{13}\right)$

$=\sin ^{-1}\left(\frac{3}{5}\right)-\sin ^{-1}\left(\frac{5}{13}\right)$

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

$=\sin ^{-1}\left(\frac{36}{65}-\frac{20}{65}\right)$

$=\sin ^{-1}\left(\frac{36-20}{65}\right)=\sin ^{-1}\left(\frac{16}{65}\right)=$ R.H.S.