Proof that $\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}$ using the unit circle. (And yes MathJax is uglier than MathML was but chrome and IE dropped support for MathML.)

Diagram used under GNU Free Documentation License.

Using the diagram above we can argue (based on the line segment $OA$ forming a right angle with $FE$ by definition of tangent line and by the triangle congruency theorem (angle-side-angle)) that the triangle formed by $A, O,$and $C$ is similar (same angles) as the triangle $A,O,$and $E$. That means there is some constant scaling factor that relates the two triangles, let's call it $k$.

Therefore we have:

$AC = kAE$

and

$kOC = OA$.

We know the length of $OA = 1$

therefore, solving for $k$ we get $k = \frac{1}{OC}$.

Now we can substitute $k$ back into $AC = kAE$ and get $AC = \frac{1}{OC}AE = \frac{AE}{OC}$.

Solving for $AE$ we get:

$AE = \frac{AC}{OC}$.

$AC$ is $\sin(\alpha)$.

$OC$ is $\cos(\alpha)$.

Therefore, $AE = \frac{\sin(\alpha)}{\cos(\alpha)}$

Since the length of $AE$ is a function of the angle in $\frac{\sin(\alpha)}{\cos(\alpha)}$ and $AE$ is on the line tangent to the unit circle at $A$ lets call it $\tan(\alpha)$.

So $\tan(\alpha)$ = $\frac{\sin(\alpha)}{\cos(\alpha)} = \frac{\text{adjacent}}{\text{opposite}}$.

QED