# Equivalence of Definitions of Tangent of Angle

## Theorem

Let $\theta$ be an angle.

The following definitions of the concept of Tangent of $\theta$ are equivalent:

### Definition from Triangle In the above right triangle, we are concerned about the angle $\theta$.

The tangent of $\angle \theta$ is defined as being $\dfrac{\text{Opposite}} {\text{Adjacent}}$.

### Definition from Circle

Consider a unit circle $C$ whose center is at the origin of a cartesian coordinate plane. Let $P$ be the point on $C$ in the first quadrant such that $\theta$ is the angle made by $OP$ with the $x$-axis.

Let a tangent line be drawn to touch $C$ at $A = \left({1, 0}\right)$.

Let $OP$ be produced to meet this tangent line at $B$.

Then the tangent of $\theta$ is defined as the length of $AB$.

## Proof

### Definition from Triangle implies Definition from Circle

Let $\tan \theta$ be defined as $\dfrac {\text{Opposite}} {\text{Adjacent}}$ in a right triangle.

Consider the triangle $\triangle OAB$.

By construction, $\angle OAB$ is a right angle.

Thus:

 $\displaystyle \tan \theta$ $=$ $\displaystyle \frac {AB} {OA}$ $\displaystyle$ $=$ $\displaystyle \frac {AB} 1$ as $OA$ is the radius of the unit circle $\displaystyle$ $=$ $\displaystyle AB$

That is:

$\tan \theta = AB$

$\Box$

### Definition from Circle implies Definition from Triangle

Let $\tan \theta$ be defined as the length of $AB$ in the triangle $\triangle OAB$.

Compare $\triangle OAB$ with $\triangle ABC$ in the diagram above.

We have that:

$\angle CAB = \angle BOA = \theta$
$\angle ABC = \angle OAB$ which is a right angle

Therefore by Triangles with Two Equal Angles are Similar it follows that $\triangle OAB$ and $\triangle ABC$ are similar.

By definition of similarity:

 $\displaystyle \frac {\text{Opposite} } {\text{Adjacent} }$ $=$ $\displaystyle \frac {BC} {AB}$ by definition $\displaystyle$ $=$ $\displaystyle \frac {AB} {OA}$ by definition of similarity $\displaystyle$ $=$ $\displaystyle AB$ as $OA$ is the radius of the unit circle $\displaystyle$ $=$ $\displaystyle \tan \theta$ by definition

That is:

$\dfrac {\text{Opposite} } {\text{Adjacent} } = \tan \theta$

$\blacksquare$