Equivalence of Definitions of Tangent of Angle

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Theorem

Let $\theta$ be an angle.

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

Definition from Triangle

SineCosine.png

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.


TangentFirstQuadrant.png


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:

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

$\blacksquare$