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

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

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?