Geometrical proof of: $\tan{(a+b)}=\frac{\tan{a}+\tan{b}}{1-\tan{a}\tan{b}}$

Question

$$\tan{(a+b)}=\frac{\tan{a}+\tan{b}}{1-\tan{a}\tan{b}}$$

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.

Answer 1

From the List of trigonometric identities on Wikipedia: