Proof of $tangent$ definition

Question

I've seen a lot of "proofs" for this - but they all somehow revolve around the definition of $tan(x)$ being $\frac{opposite}{adjacent}$. I am coming from the standpoint of being unconvinced that this is true, while looking at the following diagram:

Forgetting for a minute that $\overline{AB}$ is defined as the "tangent" - How can I geometrically prove that $\overline{AB}$ is really $\frac{opposite}{adjacent}$ or $\frac{sin}{cos}$?

I believe this would also probably not be able to avoid the fact that $\overline{OB}$ is cosecant. I likewise can't probe this (hope I can once $tan(\theta)$ is clear...

Answer 1

Use similarity. $\Delta OAC$ and $\Delta ABC$ are similar. When two triangles have congruent angles, then they must be similar. When this is the case, the ratio between the lengths of corresponding sides must be equivalent. Use $\angle O$ as the reference.

$$\Delta OAC \implies \frac{adjacent}{hypotenuse} = \frac{\cos O}{1}$$

$$\Delta ABC \implies \frac{adjacent}{hypotenuse} = \frac{\sin O}{\overline{AB}}$$

The two ratios must be equivalent.

$$\frac{\cos O}{1} = \frac{\sin O}{\overline{AB}}$$

$$\cos O\cdot \overline{AB} = \sin O \implies \overline{AB} = \frac{\sin O}{\cos O} \implies \overline{AB} = \tan O$$

Answer 2

Triangles $OAC$ and $ABC$ are similar since they have the same angles. It follows that appropriate side lengths are proportional, in particular:

$$\frac{|AB|}{|OA|} = \frac{|AC|}{|OC|} \implies \frac{|AB|}{1} = \frac{\sin x}{\cos x}\implies |AB|=\frac{\sin x}{\cos x}.$$

Answer 3

Not taking "credit" for the answer - but my own visual summary of the good answers provided: