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.