I think I would arrange the equation as:

\(\displaystyle \arcsin\left(\frac{12}{13}\right)+ \arccos\left(\frac{4}{5}\right)= \pi-\arctan\left(\frac{63}{16}\right)\)

Take the sine of both sides (apply the angle sum identity for sine on the left and the identity \(\displaystyle \sin(\pi-\theta)=\sin(\theta)\) on the right):

\(\displaystyle \sin\left(\arcsin\left(\frac{12}{13}\right)\right) \cos\left(\arccos\left(\frac{4}{5}\right)\right)+ \cos\left(\arcsin\left(\frac{12}{13}\right)\right) \sin\left(\arccos\left(\frac{4}{5}\right)\right)= \sin\left(\arctan\left(\frac{63}{16}\right)\right)\)

\(\displaystyle \frac{12}{13}\cdot\frac{4}{5}+ \frac{5}{13}\cdot\frac{3}{5}=\frac{63}{65}\)

\(\displaystyle \frac{48+15}{65}=\frac{63}{65}\)

\(\displaystyle \frac{63}{65}=\frac{63}{65}\quad\checkmark\)

This argument is not valid.

To see this, start with the following equation which is obviously not correct

\(\displaystyle \sin ^{-1}\frac{12}{13}+\cos ^{-1}\frac{4}{5}=\tan ^{-1}\frac{63}{16}\)

and apply $sin$ to both sides to get

\(\displaystyle \frac{63}{65}=\frac{63}{65}\)

so taking sin of both sides and getting equal answers does not prove the original assertion