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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 $\cot(\theta)$.

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    $\begingroup$ It’s all about similar triangles. $\endgroup$ Jan 8 at 19:29
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    $\begingroup$ $$\dfrac\sin\cos=\dfrac\tan1$$ $\endgroup$
    – user65203
    Jan 8 at 19:41
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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 $\triangle OCA$.

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  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.
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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.

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