$\begin{align}  & \text{Let }{{\sin }^{-1}}\left( \frac{3}{5} \right)=\theta  \\ & \sin \theta =\frac{3}{5},\text{ cos}\theta \text{=}\frac{4}{5},\text{ }\tan \theta =\frac{3}{4} \\ & \theta ={{\sin }^{-1}}\left( \frac{3}{5} \right)={{\cos }^{-1}}\frac{4}{5}={{\tan }^{-1}}\left( \frac{3}{4} \right) \\ & \text{Let }{{\cos }^{-1}}\left( \frac{63}{65} \right)=\alpha  \\ & \cos \alpha =\frac{63}{65},\text{ }\sin \alpha =\frac{16}{65},\text{ }\tan \alpha =\frac{16}{63} \\ & \alpha ={{\sin }^{-1}}\left( \frac{16}{65} \right)={{\cos }^{-1}}\left( \frac{63}{65} \right)={{\tan }^{-1}}\left( \frac{16}{63} \right) \\ & \text{From the L}\text{.H}\text{.S} \\ & {{\sin }^{-1}}(\tfrac{3}{5})-{{\cos }^{-1}}(\tfrac{63}{65})={{\tan }^{-1}}(\tfrac{3}{4})-{{\tan }^{-1}}(\tfrac{16}{63}) \\ & {{\tan }^{-1}}(\tfrac{3}{4})-{{\tan }^{-1}}(\tfrac{16}{63})={{\tan }^{-1}}\tan \left[ {{\tan }^{-1}}(\tfrac{3}{4})-{{\tan }^{-1}}(\tfrac{16}{63}) \right] \\ & ={{\tan }^{-1}}\left[ \frac{\tan {{\tan }^{-1}}(\tfrac{3}{4})-\tan {{\tan }^{-1}}(\tfrac{16}{63})}{1+\tan {{\tan }^{-1}}(\tfrac{3}{4})\tan {{\tan }^{-1}}(\tfrac{16}{63})} \right] \\ & ={{\tan }^{-1}}\left[ \frac{\tfrac{3}{4}-\tfrac{16}{63}}{1+(\tfrac{3}{4})(\tfrac{16}{63})} \right] \\ & ={{\tan }^{-1}}\left[ \frac{{125}/{252}\;}{{25}/{21}\;} \right] \\ & ={{\tan }^{-1}}\left( \frac{125}{252}\times \frac{21}{25} \right)={{\tan }^{-1}}\left( \frac{5}{21} \right) \\\end{align}$