Prove $\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{56}{65}$

(A) Let $\sin ^{-1} \frac{3}{5}=x$. Then,$\sin x=\frac{3}{5} \Rightarrow \cos x=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$.
$\therefore \tan x=\frac{3}{4} \Rightarrow x=\tan ^{-1} \frac{3}{4}$.
$\therefore \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{3}{4} \dots(1)$.
Now,let $\cos ^{-1} \frac{12}{13}=y$. Then,$\cos y=\frac{12}{13} \Rightarrow \sin y=\sqrt{1-\left(\frac{12}{13}\right)^{2}}=\sqrt{\frac{25}{169}}=\frac{5}{13}$.
$\therefore \tan y=\frac{5}{12} \Rightarrow y=\tan ^{-1} \frac{5}{12}$.
$\therefore \cos ^{-1} \frac{12}{13}=\tan ^{-1} \frac{5}{12} \dots(2)$.
Now,consider the $L$.$H$.$S$: $\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}$.
Using equations $(1)$ and $(2)$,we get $\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{3}{4}$.
Using the formula $\tan ^{-1} A + \tan ^{-1} B = \tan ^{-1} \left( \frac{A+B}{1-AB} \right)$:
$= \tan ^{-1} \left( \frac{\frac{5}{12}+\frac{3}{4}}{1-\left(\frac{5}{12} \cdot \frac{3}{4}\right)} \right) = \tan ^{-1} \left( \frac{\frac{20+36}{48}}{\frac{48-15}{48}} \right) = \tan ^{-1} \left( \frac{56}{33} \right)$.
To convert $\tan ^{-1} \frac{56}{33}$ to $\sin ^{-1}$,let $\tan ^{-1} \frac{56}{33} = z$. Then $\tan z = \frac{56}{33}$.
Using the identity $\sin z = \frac{\tan z}{\sqrt{1+\tan^2 z}} = \frac{56/33}{\sqrt{1+(56/33)^2}} = \frac{56/33}{\sqrt{(1089+3136)/1089}} = \frac{56/33}{65/33} = \frac{56}{65}$.
Thus,$\tan ^{-1} \frac{56}{33} = \sin ^{-1} \frac{56}{65} = R.H.S$.