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 $AB$.
Hence in the first quadrant, the cotangent is positive.
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.
From Parallelism implies Equal Alternate Angles:
- $\angle OBA = \theta$
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.
From Parallelism implies Equal Alternate Angles:
- $\angle OBA = \theta$
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:
- $\dfrac {\text{Adjacent} } {\text{Opposite} } = \cot \theta$
$\blacksquare$