Originally Posted by

**MarkFL** 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):

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

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

$\displaystyle \frac{48+15}{65}=\frac{63}{65}$

$\displaystyle \frac{63}{65}=\frac{63}{65}\quad\checkmark$