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

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.

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From the List of trigonometric identities on Wikipedia:

Geometric proof of tan(a+b)

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