Graphs of Tangent, Cotangent, Secant, and Cosecant

13 min read

Point a phone clinometer at the top of a tree and it reports a height. Tilt the line of sight closer to straight up and that height races toward huge numbers, then refuses to answer at exactly vertical. That runaway is the tangent function. Tangent, cotangent, secant, and cosecant are each a ratio or a reciprocal of sine and cosine, so each one falls apart into repeating branches split by vertical lines. Place those lines and the rest of the graph fills in from a wave you already know.

By the end of this page you can:

• State the period of each of the four functions and say why tangent and cotangent repeat twice as fast as the others.
• Locate the vertical asymptotes of each function from the zeros of sine or cosine.
• Read the stretching factor and the period from $y = a\tan(k x)$.
• Find two consecutive asymptotes and the phase shift of a transformed tangent or cotangent.

The floor. The asymptotes sit exactly where the denominator (sine or cosine) is zero; tangent and cotangent have period $\pi$, and secant and cosecant have period $2\pi$. Placing those vertical lines and repeating at the right period is most of the credit, even before the curve between the lines is drawn.

Quick Reference

Before You Start

Check each box you can do from memory. A box you cannot check yet points to a quick refresher, not a grade.

• I can sketch one period of $y = \sin x$ and $y = \cos x$ and mark where each one equals $0$, $1$, and $-1$. (Graphs of Sine and Cosine)
• I can list the inputs in $[0, 2\pi]$ where cosine is zero and where sine is zero.
• I can say what happens to a fraction whose denominator approaches $0$ while its numerator stays away from $0$. (Signs and Domains of the Trig Functions)

A 60-second self-check. Pick an answer, then reveal the reasoning.

The Idea Behind These Four Graphs

Start with cosine, which you can already draw. Tangent is $\tan x = \dfrac{\sin x}{\cos x}$, so the cosine in the denominator controls the graph. On a copy of the cosine wave over $[-2\pi, 2\pi]$, mark every input where $\cos x = 0$. Ask what a fraction does when its denominator hits $0$ while the numerator does not, and predict what tangent does near each mark before reading on.

Each of the four functions is built from sine and cosine, so each graph is read off a wave you know.

Two ideas carry every graph here. A ratio blows up where its denominator is zero, which is the vertical asymptote. A reciprocal is large where the original is small and small where the original is large, which gives the U-shaped branches of secant and cosecant riding above and below the cosine and sine waves.

One contrast is worth holding onto. Tangent and cotangent repeat every $\pi$, half the period of sine and cosine, because the ratio $\sin/\cos$ completes its sign pattern in half the distance. Secant and cosecant keep the period $2\pi$ of the functions they invert, because a reciprocal does not change where a function repeats.

Prerequisite Hub

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Builds on:

Unlocks (what this section feeds into next):

Cross-course prerequisites: none.

Official Definitions

Each statement below is the canonical fact to put on a study sheet. The period and asymptote facts are stated in Stewart 7e, Section 5.4; the reciprocal and quotient identities are the standard definitions of the other four functions.

Period of tangent and cotangent

The tangent and cotangent functions have period $\pi$. The graph of $y = \tan x$ repeats on every interval of length $\pi$, and the same holds for $y = \cot x$.

Stewart 7e, Section 5.4 More Trigonometric Graphs.

Period of secant and cosecant

The secant and cosecant functions have period $2\pi$. The graph of $y = \sec x$ repeats on every interval of length $2\pi$, and the same holds for $y = \csc x$.

Stewart 7e, Section 5.4 More Trigonometric Graphs.

Transformations of tangent and cotangent

For $y = a\tan(kx)$ and $y = a\cot(kx)$ with $k > 0$, the period is $\dfrac{\pi}{k}$. For $y = a\tan(k(x - b))$ the graph is shifted horizontally by $b$, the phase shift.

Stewart 7e, Section 5.4 More Trigonometric Graphs.

Vertical asymptotes from a zero denominator

Tangent and secant have vertical asymptotes where $\cos x = 0$, that is at $x = \dfrac{\pi}{2} + n\pi$ for integers $n$. Cotangent and cosecant have vertical asymptotes where $\sin x = 0$, that is at $x = n\pi$ for integers $n$.

Stewart 7e, Section 5.4 More Trigonometric Graphs.

OpenStax states the transformed tangent asymptotes directly, and the quote is exact:

“For $y=A\tan(Bx)$ ... vertical asymptotes occur at $x=\dfrac{\pi}{2|B|}+\dfrac{\pi}{|B|}k$, where $k$ is an integer.”

OpenStax Precalculus 2e, Section 6.2 Graphs of the Other Trigonometric Functions.

At a Glance

Sine and cosine stay between $-1$ and $1$. Tangent and cotangent reach every real value, and secant and cosecant skip the open band between $-1$ and $1$.

Worked Examples

Example 1: Period and asymptotes of a stretched tangent

Find the stretching factor, the period, and two consecutive vertical asymptotes for $y = 3\tan(2x)$.

Predict first. The coefficient $2$ on $x$ compresses the graph horizontally, so the period should come in shorter than the standard $\pi$.

Solution.

Match the form $y = a\tan(kx)$, so $a = 3$ and $k = 2$.

The stretching factor is $|a| = 3$.

The period is \[ P = \frac{\pi}{k} = \frac{\pi}{2}. \]

The asymptotes occur where the inside $2x$ equals an odd multiple of $\dfrac{\pi}{2}$, that is at \[ x = \frac{\pi}{2 \cdot 2} + \frac{\pi}{2}n = \frac{\pi}{4} + \frac{\pi}{2}n, \quad n \text{ an integer}. \]

Taking $n = 0$ and $n = 1$ gives two consecutive asymptotes at $x = \dfrac{\pi}{4}$ and $x = \dfrac{3\pi}{4}$.

Check the prediction. The period $\dfrac{\pi}{2}$ is shorter than $\pi$, matching the compression from $k = 2$. The gap between the asymptotes is $\dfrac{3\pi}{4} - \dfrac{\pi}{4} = \dfrac{\pi}{2}$, one full period, which is the spacing tangent asymptotes must have.

Example 2: Vertical asymptotes of a cosecant graph

State the period and the vertical asymptotes of $y = \csc x$ on $[0, 2\pi]$.

Predict first. Cosecant is the reciprocal of sine, so its asymptotes belong at the zeros of sine.

Solution.

Cosecant is $\csc x = \dfrac{1}{\sin x}$, so a vertical asymptote occurs wherever $\sin x = 0$.

On $[0, 2\pi]$, sine is zero at $x = 0$, $x = \pi$, and $x = 2\pi$.

The period of cosecant is $2\pi$.

So on $[0, 2\pi]$ the vertical asymptotes are the lines $x = 0$, $x = \pi$, and $x = 2\pi$.

Check the prediction. Between consecutive asymptotes, sine has a single hump that reaches $1$ or $-1$. The reciprocal of that hump is a U-shaped branch whose turning point sits at $1$ or $-1$, matching the cosecant range $(-\infty, -1] \cup [1, \infty)$.

Example 3: Building a cotangent period by hand

Plot one full period of $y = \cot x$ using only the zeros and signs of sine and cosine.

Predict first. Cotangent is $\dfrac{\cos x}{\sin x}$. Predict where the graph touches the horizontal axis (numerator zero) and where it has asymptotes (denominator zero) before plotting.

Solution.

Write $\cot x = \dfrac{\cos x}{\sin x}$.

The graph crosses the horizontal axis where the numerator is zero, that is where $\cos x = 0$. On $(0, \pi)$ this happens at $x = \dfrac{\pi}{2}$, so $\left(\dfrac{\pi}{2}, 0\right)$ is on the graph.

The graph has asymptotes where the denominator is zero, that is where $\sin x = 0$. The endpoints $x = 0$ and $x = \pi$ are asymptotes.

Track the sign across one period:

One period of cotangent falls from $+\infty$ through $1$, $0$, and $-1$ to $-\infty$, with asymptotes at $x = 0$ and $x = \pi$ and one axis crossing at $x = \dfrac{\pi}{2}$.

Check the prediction. The decreasing run from $+\infty$ to $-\infty$ over a width of $\pi$ matches the cotangent period of $\pi$ and its range of all real numbers.

Example 4: Period, asymptotes, and phase shift together

For $y = 2\tan\!\left(x - \dfrac{\pi}{4}\right)$, find the period, the phase shift, and the two asymptotes nearest the phase shift.

Predict first. The coefficient on $x$ inside is $1$, so the period should stay at $\pi$. Subtracting $\dfrac{\pi}{4}$ slides the graph to the right.

Solution.

Match $y = a\tan(k(x - b))$, so $a = 2$, $k = 1$, and $b = \dfrac{\pi}{4}$.

The period is \[ P = \frac{\pi}{k} = \frac{\pi}{1} = \pi. \]

The phase shift is $b = \dfrac{\pi}{4}$ to the right.

The basic tangent $y = \tan x$ has asymptotes at $x = -\dfrac{\pi}{2}$ and $x = \dfrac{\pi}{2}$ around its central branch. Shifting right by $\dfrac{\pi}{4}$ moves these to \[ x = -\frac{\pi}{2} + \frac{\pi}{4} = -\frac{\pi}{4} \quad\text{and}\quad x = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}. \]

Check the prediction. The period is $\pi$, as predicted from $k = 1$. The two asymptotes $-\dfrac{\pi}{4}$ and $\dfrac{3\pi}{4}$ are $\pi$ apart, one full period, and the central branch is centered on the phase shift $\dfrac{\pi}{4}$, exactly halfway between them. That centering confirms the shift was applied correctly.

Common Misconceptions

Practice Problems

Check Yourself

Close the notes and answer each from memory, then reveal it. Pulling an idea back from memory is one of the strongest ways to make it stick.

Go Deeper (optional)

None of this is needed to sketch the four graphs or to pass the section. It is here for readers who want to see where these functions show up.

Tangent is the function behind angle-of-elevation measurement. A surveyor standing a known distance $d$ from a tower reads the angle $\theta$ up to the top, and the height is $d\tan\theta$. As $\theta$ approaches $\dfrac{\pi}{2}$, $\tan\theta$ runs toward $+\infty$, which is the graph’s asymptote made physical: a nearly vertical sight line forces an enormous computed height for any fixed base distance. The same runaway is why a camera tilted toward straight up cannot frame the top of a tall building from up close.

Secant and cosecant appear in distances along a slanted path. On a ramp that rises at angle $\theta$, the slope length for a fixed horizontal run is proportional to $\sec\theta$, and for a fixed vertical rise it is proportional to $\csc\theta$. Because $\sec\theta \ge 1$ always, the slanted distance is never shorter than the straight run, which is the geometric meaning of the range $(-\infty, -1] \cup [1, \infty)$.

Mastery Checklist

Novice (Level 1-2):

• I can state the period of each of tangent, cotangent, secant, and cosecant.
• I can name which sine or cosine zeros produce the asymptotes of each function.
• I can locate the vertical asymptotes of a basic secant or cosecant graph on $[0, 2\pi]$.

Competent (Level 3-4):

• I can read the stretching factor $|a|$ and period $\dfrac{\pi}{k}$ from $y = a\tan(kx)$.
• I can find two consecutive asymptotes of $y = a\tan(kx)$ from the inside-equals-odd-multiple-of-pi-over-2 condition.
• I can find the phase shift of $y = a\tan(k(x - b))$ and plot one period of cotangent using the zeros and signs of sine and cosine.

Proficient (Level 5):

• I can justify the range of secant and cosecant from the bound on sine and cosine.
• I can decide whether a reciprocal-trig equation has a solution by reasoning about range.
• I can explain why a vertical asymptote marks an undefined input rather than a wall the graph stops at.

Connections

Looking back:

• Graphs of Sine and Cosine supply the waves these four functions are read from.
• Signs and Domains of the Trig Functions fix the inputs where each function is undefined, which become the asymptotes.

Looking ahead:

• Inverse Trigonometric Functions restrict each of these graphs to a one-to-one piece, which is what makes an inverse possible.

Real-world connections:

• Tangent measures the height-to-distance ratio that an angle of elevation produces, which is why it grows without bound as the line of sight approaches vertical.
• Secant and cosecant give distances along a slanted line, where a small angle produces a large reciprocal length.

Resources

Last updated: 2026-06-24