I put this proof together because, surprisingly, I couldn’t find a simple geometric explanation of the identity. In school this sort of thing always bothered me and made me resent math as a kid. Most resources either jump straight to coordinate definitions or algebraic identities, but they skip over the kind of intuitive, visual argument that makes the identity feel true. This version sticks with pure geometry, using triangle similarity and the unit circle, to show where the tangent function really comes from.

Proof that \( \tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} \) using the unit circle

Unit circle diagram

Diagram used under GNU Free Documentation License.

Using the diagram above, we argue (based on the line segment \( OA \) forming a right angle with \( FE \) by the definition of a tangent line, and by the triangle congruence theorem (angle-side-angle)) that the triangle formed by \( \triangle AOC \) is similar to \( \triangle AOE \) — since they share angle \( \alpha \) and right angles.

Because the triangles are similar, there exists a constant scaling factor \( k \) such that:

\[ AC = k \cdot AE \quad \text{and} \quad k \cdot OC = OA \]

Since the unit circle has radius 1, we know \( OA = 1 \). Solving for \( k \), we find: \[ k = \frac{1}{OC} \]

Substituting into the first equation: \[ AC = \frac{1}{OC} \cdot AE \Rightarrow AE = \frac{AC}{OC} \]

From the diagram: \[ AC = \sin(\alpha), \quad OC = \cos(\alpha) \]

Thus, \[ AE = \frac{\sin(\alpha)}{\cos(\alpha)} \]

Because \( AE \) lies along the tangent to the unit circle at point \( A \), and its length is a function of the angle \( \alpha \), we define it as \( \tan(\alpha) \). Therefore: \[ \tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} \]

Q.E.D.