# How does the length of the tangent in this picture relate to $\sin(x)/\cos(x)$?

I understand the justificaton behind the length of $$\sin(\theta)$$ and $$\cos(\theta)$$. But I don't understand why AE is the length of $$\tan(\theta)$$, I understand that it satisfies the pythagorean identities but how exactly is that length related to $$\frac{\sin(x)}{\cos(x)}$$? I also don't understand the lengths of the other trigonometric functions $$\sec(\theta)$$, $$\csc(\theta)$$ and $$\cot(\theta)$$.

• It’s all about similar triangles. Jan 8, 2021 at 19:29
• $$\dfrac\sin\cos=\dfrac\tan1$$
– user65203
Jan 8, 2021 at 19:41

Right-angled triangles with one angle equal to $$\theta$$ include $$\triangle OAE,\,FAO$$, each with $$\theta$$ (the right-angle) at the vertex listed first (second). Now use similarity to $$\triangle OCA$$.

1. $$\angle OAE = \frac{\pi}{2} \ (= 90^\circ)$$,
2. $$\angle CAE = \angle OAE - \angle OAC = \frac{\pi}{2} - \left(\frac{\pi}{2} - \angle COA\right) = \theta$$,
3. So now $$\Delta CAE$$ and $$\Delta COA$$ share two of the same angles and so must be similar,
4. Hence the ratio of their sides must be equal, and the result follows.

Consider the intersection point $$T$$ of line $$OA$$ with the tangent to the circle at $$D$$. By definition the length $$DT$$ is equal to $$\tan \theta$$ since we have a unit circle.

Now the right triangles $$OTD$$ and $$OAE$$ are isometric, by the second congruence case, and therefore, the corresponding edges $$DT$$ and $$AE$$ have the same length.