$$\tan{(a+b)}=\frac{\tan{a}+\tan{b}}{1-\tan{a}\tan{b}}$$
The way my textbook develops this identity is by first deriving both $\sin{(a+b)}$ and $\cos{(a+b)}$ geometrically and then simply expressing $\tan{(a+b)}$ as $\frac{\sin{(a+b)}}{\cos{(a+b)}}$.
But is there a way to directly derive this identity geometrically?
I did find this question here already but it wasn't answered.