Equivalence of Definitions of Secant of Angle
Theorem
Let $\theta$ be an angle.
The following definitions of the concept of secant are equivalent:
Definition from Triangle
In the above right triangle, we are concerned about the angle $\theta$.
The secant of $\angle \theta$ is defined as being $\dfrac{\text{Hypotenuse}} {\text{Adjacent}}$.
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 $A = \tuple {1, 0}$.
Let $OP$ be produced to meet this tangent line at $B$.
Then the secant of $\theta$ is defined as the length of $OB$.
Hence in the first quadrant, the secant is positive.
Proof
Definition from Triangle implies Definition from Circle
Let $\sec \theta$ be defined as $\dfrac {\text{Hypotenuse}} {\text{Adjacent}}$ in a right triangle.
Consider the triangle $\triangle OAB$.
By construction, $\angle OAB$ is a right angle.
Thus:
| \(\ds \sec \theta\) | \(=\) | \(\ds \frac {OB} {OA}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {OB} 1\) | as $OA$ is the radius of the unit circle | |||||||||||
| \(\ds \) | \(=\) | \(\ds OB\) |
That is:
- $\sec \theta = OB$
$\Box$
Definition from Circle implies Definition from Triangle
Let $\sec \theta$ be defined as the length of $OB$ 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:
| \(\ds \frac {\text{Hypotenuse} } {\text{Adjacent} }\) | \(=\) | \(\ds \frac {AC} {AB}\) | by definition | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {OB} {OA}\) | by definition of similarity | |||||||||||
| \(\ds \) | \(=\) | \(\ds OB\) | as $OA$ is the radius of the unit circle | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sec \theta\) | by definition |
That is:
- $\dfrac {\text{Hypotenuse} } {\text{Adjacent} } = \sec \theta$
$\blacksquare$