The secant, cosecant, and cotangent functions

AP Precalculus· 20 min read

1. Core Definitions of Reciprocal Trigonometric Functions★★☆☆☆⏱ 5 min

📘 Definition

Reciprocal Trigonometric Functions

$sec θ$ , $csc θ$ , $cot θ$

The three trigonometric functions defined as the multiplicative inverses of cosine, sine, and tangent respectively, for all inputs where the base function is non-zero.

\sec \theta = \frac{1}{\cos \theta}, \quad \csc \theta = \frac{1}{\sin \theta}, \quad \cot \theta = \frac{\cos \theta}{\sin \theta}

Note that cotangent can be written both as $1/ tan θ$ and the ratio of cosine to sine; the second form avoids undefined values when tangent itself is undefined, which simplifies asymptote calculations.

📐 Worked Example

Evaluate $sec (6 0^{\circ})$ , $csc (3 π /2)$ , and $cot (5 π /4)$ using unit circle values.

1
First retrieve the base sine and cosine values for each angle:
$\cos 60^\circ = 0.5, \quad \sin\left(\frac{3\pi}{2}\right) = -1, \quad \cos\left(\frac{5\pi}{4}\right) = \sin\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}$
2
Apply the reciprocal definitions:
$\sec 60^\circ = \frac{1}{0.5} = 2, \quad \csc\left(\frac{3\pi}{2}\right) = \frac{1}{-1} = -1, \quad \cot\left(\frac{5\pi}{4}\right) = \frac{-\sqrt{2}/2}{-\sqrt{2}/2} = 1$

Exam tip:

The AP exam almost never asks you to compute these values directly, but you will need them as intermediate steps for larger problems.

2. Domain, Range, and Asymptote Properties★★★☆☆⏱ 6 min

Since each reciprocal function is undefined when its base sine or cosine is zero, we can derive vertical asymptote locations directly from the zeros of the base function.

Function	Undefined When	Vertical Asymptotes (radians)	Domain	Range
$sec θ$	$cos θ = 0$	$θ = \frac{π}{2} + k π$ for all integers $k$	All real numbers except $\frac{π}{2} + k π$	$(- \infty, - 1] \cup [1, \infty)$
$csc θ$	$sin θ = 0$	$θ = k π$ for all integers $k$	All real numbers except $k π$	$(- \infty, - 1] \cup [1, \infty)$
$cot θ$	$sin θ = 0$	$θ = k π$ for all integers $k$	All real numbers except $k π$	All real numbers

✓ Quick check

Test your understanding of domain rules:

Which of the following angles is not in the domain of $csc θ$ ?
- A. $π /3$
- B. $π /2$
- C. $π$
- D. $3 π /2$
Reveal answer
C —
$sin π = 0$ , so $csc π = 1/0$ which is undefined, so $π$ is excluded from the domain of cosecant.

3. Graphs of Secant, Cosecant, and Cotangent★★★☆☆⏱ 6 min

To sketch a secant or cosecant graph, first draw the corresponding cosine or sine wave, then draw vertical asymptotes at every zero of that wave, then draw U-shaped reciprocal curves between each pair of asymptotes touching the peaks and troughs of the base sine/cosine graph.

🔬 Derivation

Goal:

Derive the period of $cot θ$

Starting from:

$tan θ$ has period $π$ , and $cot θ = - tan (θ - π /2)$

1
A horizontal shift and vertical reflection cannot change the period of a periodic function
2
Since tangent repeats every $π$ radians, cotangent must also repeat every $π$ radians

Result:

The period of untransformed $cot θ$ , $sec θ$ , and $csc θ$ is $π$ , $2 π$ , and $2 π$ respectively.

📐 Worked Example

Find the period and locations of two consecutive vertical asymptotes for $f (x) = 2 csc (3 x)$ .

1
First, identify the transformation applied to the base cosecant function. The horizontal scaling factor is 3, so the period is:
$T = \frac{2\pi}{3}$
2
Base cosecant has asymptotes at $0$ and $π$ , so scale these by $1/3$ to get the new asymptote locations:
$x=0, \quad x = \frac{\pi}{3}$

4. Reciprocal Pythagorean Identities★★★★☆⏱ 3 min

🔬 Derivation

Goal:

Derive the Pythagorean identity for secant and tangent

Starting from:

Core Pythagorean identity $sin^{2} θ + cos^{2} θ = 1$

1
Divide every term in the identity by $cos^{2} θ$
2
Simplify each term: $\frac{s i n ^{2} θ}{c o s ^{2} θ} = tan^{2} θ$ , $\frac{c o s ^{2} θ}{c o s ^{2} θ} = 1$ , $\frac{1}{c o s ^{2} θ} = sec^{2} θ$

Result:

The resulting identity is $1 + tan^{2} θ = sec^{2} θ$

1 + \cot^2 \theta = \csc^2 \theta

5. Common Pitfalls

Wrong move:

Defining cotangent as $1/ tan θ$ and then incorrectly evaluating it at $θ = π /2$ where tangent is undefined

Why:

At $θ = π /2$ , $tan θ$ is undefined, but $cot θ = cos θ / sin θ = 0$ , so the reciprocal of tangent form breaks here

Correct move:

Always use $cot θ = cos θ / sin θ$ for all calculations to avoid missing valid defined values.

Wrong move:

Thinking the range of secant includes values between -1 and 1

Why:

Students confuse the range of cosine (which is [-1,1]) with its reciprocal

Correct move:

Remember that if $∣ x ∣ \leq 1$ , then $∣1/ x ∣ \geq 1$ for non-zero $x$ .

Wrong move:

Assigning a period of $2 π$ to the cotangent function

Why:

Students assume all trigonometric functions have period $2 π$

Correct move:

Memorize that only tangent and cotangent have base period $π$ , the other four trig functions have base period $2 π$ .

Wrong move:

Drawing secant graphs that cross the x-axis

Why:

Secant is 1/cosine, which can never equal zero, since the numerator is always 1

Correct move:

Reciprocal functions of sine and cosine never have x-intercepts, only vertical asymptotes.

Wrong move:

Forgetting to scale asymptote locations when applying horizontal stretches/compressions

Why:

Students leave asymptotes at the base function positions instead of adjusting for the B value in $A sec (B x)$

Correct move:

Multiply all base asymptote x-values by $1/ B$ to get the new positions after horizontal scaling.

6. Quick Reference Cheatsheet

Function	Reciprocal Definition	Base Period	Asymptote Locations	Range
$sec θ$	$1/ cos θ$	$2 π$	$π /2 + k π$	$(- \infty, - 1] \cup [1, \infty)$
$csc θ$	$1/ sin θ$	$2 π$	$k π$	$(- \infty, - 1] \cup [1, \infty)$
$cot θ$	$cos θ / sin θ$	$π$	$k π$	$R$
Identity 1	$1 + tan^{2} θ = sec^{2} θ$
Identity 2	$1 + cot^{2} θ = csc^{2} θ$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2024 · 1
Reciprocal identity evaluation
2023 · 2
Secant graph transformation task

What's Next

Mastering these reciprocal trigonometric functions is a critical prerequisite for upcoming AP Precalculus topics, including solving trigonometric equations, verifying trigonometric identities, and working with polar coordinates. You will use secant, cosecant, and cotangent regularly in both the multiple choice and free response sections of the AP exam, especially when working with right triangle applications and periodic function modeling. These functions also form the foundation for calculus topics like derivatives of trigonometric functions that you will encounter in future courses.

→Trigonometric Identity Verification