Question

${\sin }^{-1}\frac{3}{5}+{\cos }^{-1}\frac{12}{13}={\cos }^{-1}\frac{33}{65}$

Answer 1

${\sin }^{-1}\frac{3}{5}+{\cos }^{-1}\frac{12}{13}={\cos }^{-1}\frac{33}{65}$

$\Rightarrow {\cos }^{-1}\frac{4}{5}+{\cos }^{-1}\left(\frac{12}{13}\right)={\cos }^{-1}\left(\frac{33}{65}\right)$

$a={\cos }^{-1}\left(\frac{4}{5}\right)b={\cos }^{-1}\left(\frac{12}{13}\right)$

$\cos a=\frac{4}{5}cosb=\frac{12}{13}$

$\sin a=\frac{3}{5}sinb=\frac{5}{13}$

$\cos (a+b)=\cos asinb-sinasinb$

$=\frac{4}{5}\times \frac{12}{13}-\frac{3}{5}\times \frac{5}{13}$

$=\frac{48}{65}-\frac{3}{13}$

$=\frac{48-3\left(5\right)}{65}=\frac{33}{65}$

$\cos (a+b)=\frac{33}{65}$

$a+b={\cos }^{-1}\left(\frac{33}{65}\right)$

$=RHS$