# Equivalence of Definitions of Cotangent of Angle

## Theorem

Let $\theta$ be an angle.

The following definitions of the concept of Cotangent of Angle are equivalent:

### Definition from Triangle

In the above right triangle, we are concerned about the angle $\theta$.

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

### Definition from Circle

Consider a unit circle $C$ whose center is at the origin of a cartesian 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 $A = \tuple {0, 1}$.

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

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

## Proof

### Definition from Triangle implies Definition from Circle

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

Consider the triangle $\triangle OAB$.

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

$\angle OBA = \theta$

Thus:

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

That is:

$\cot \theta = AB$

$\Box$

### Definition from Circle implies Definition from Triangle

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

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

$\angle OBA = \theta$

We have that:

$\angle CAB = \angle OBA = \theta$
$\angle ABC = \angle OAB$ 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:

 $\ds \frac {\text{Adjacent} } {\text{Opposite} }$ $=$ $\ds \frac {AB} {BC}$ by definition $\ds$ $=$ $\ds \frac {AB} {OA}$ Definition of Similar Triangles $\ds$ $=$ $\ds AB$ $OA$ is Radius of the Unit Circle $\ds$ $=$ $\ds \cot \theta$ by definition

That is:

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

$\blacksquare$