Precalculus

Secant, cosecant and cotangent on the unit circle

Sine and cosine have a value at every angle. Tangent, secant, cosecant and cotangent each divide by a coordinate, and a coordinate can be zero. Which angles are left without a value, and which values can these functions never take?

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Trig functions on the unit circlePrecalculus Secant, cosecant and cotangent ratiosPrecalculus Domain and rangeGrade 9

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Exact values on the unit circlePrecalculus Graphing tangent and the reciprocal functionsPrecalculus Reciprocal and quotient identitiesPrecalculus

Where you are 8 in The unit circle

Trigonometry7

The unit circle8

▴ 1 above Periodicity and the periodicity identities Reference angles

Secant, cosecant and cotangent on the unit circle

Signs of the trig functions by quadrant Sine, cosine and tangent as functions of a real number ▾ Show all 8 (2 more)

Why this exists

In a right triangle the secant, cosecant and cotangent are the cosine, sine and tangent turned upside down. A triangle never has a side of length 0, so the triangle versions are never in danger. Once cosine and sine become coordinates on the unit circle, every angle has a terminal point, and that point may sit on an axis with a coordinate of 0. Turning a 0 upside down is impossible.

So the circle raises a question the triangle never asked: at which angles does each function have no value, and which outputs can it produce? The answers are the domain and range of the six functions. They decide where graphs break and which equations have no solution.

Definition

Let \theta be an angle in standard position with terminal point (x, y) on the unit circle. The six trigonometric functions of \theta are

\begin{aligned} \sin\theta &= y & \cos\theta &= x \\ \csc\theta &= \frac{1}{y} & \sec\theta &= \frac{1}{x} \\ \cot\theta &= \frac{x}{y} & \tan\theta &= \frac{y}{x} \end{aligned}

Sine and cosine always exist. Cosecant and cotangent divide by y, so they need y \ne 0; secant and tangent divide by x, so they need x \ne 0.

In words: the secant is one over the cosine, the cosecant is one over the sine, and the cotangent is the cosine divided by the sine. When its condition fails, the function is undefined at \theta: it has no value there, not the value 0. For acute angles these match the triangle ratios.

Intuition

Put a searchlight at the origin and a long straight wall along the line x = 1, one unit from the light. The beam leaves at angle \theta, measured from the perpendicular to the wall (the positive x-axis). The beam lies on the line through the origin with slope \tan\theta, so it reaches the wall at the point (1, \tan\theta): the tangent is how high the spot is on the wall. The beam’s length from the light to that spot is \sec\theta: the secant is how far the light travels.

Figure 1A light at the origin and a wall along x = 1. The beam reaches the wall at height \tan\theta, and its length to the wall is \sec\theta.

Now turn the beam toward 90^\circ. The spot races up the wall and the beam gets longer and longer. At exactly 90^\circ the beam points straight along the y-axis, parallel to the wall, and never reaches it. There is no spot and no length, so there is no \tan 90^\circ and no \sec 90^\circ. The same happens at 270^\circ, pointing straight down. In coordinates, those are the two angles whose terminal point has x = 0.

Figure 2As the beam turns toward 90^\circ its spot climbs the wall without limit. At 90^\circ the beam is parallel to the wall, so \tan 90^\circ and \sec 90^\circ do not exist.

A second wall along y = 1 does the same for the other pair: the beam meets it at (\cot\theta, 1), at distance \csc\theta. A beam along the x-axis (y = 0) is parallel to it, so cotangent and cosecant fail at 0^\circ and 180^\circ.

Figure 3A wall along y = 1 does the same job for the other pair. The beam meets it at (\cot\theta, 1), at distance \csc\theta from the light.

The picture shows the ranges too. The wall is one unit away, so no beam reaching it is shorter than 1: secant and cosecant never lie strictly between -1 and 1. The spot can be anywhere on the wall, so tangent and cotangent take every real value.

Results

Where each function is undefined. With k standing for any integer,

Function	Fails when	Domain
\sin, \cos	never	all real \theta
\tan, \sec	x = 0	\displaystyle \theta \ne \frac{\pi}{2} + k\pi
\cot, \csc	y = 0	\theta \ne k\pi

In degrees the gaps are 90^\circ + k \cdot 180^\circ and k \cdot 180^\circ.

Ranges.

Function	Range
\sin\theta, \cos\theta	[-1, 1]
\tan\theta, \cot\theta	all real numbers
\sec\theta, \csc\theta	(-\infty, -1] \cup [1, \infty)

Reciprocals and a quotient. Wherever both sides are defined,

\begin{gathered} \sec\theta = \frac{1}{\cos\theta} \qquad \csc\theta = \frac{1}{\sin\theta} \\ \cot\theta = \frac{\cos\theta}{\sin\theta} \end{gathered}

and \displaystyle \cot\theta = \frac{1}{\tan\theta} wherever \tan\theta is defined and not 0.

Distances. The line through the terminal side meets x = 1 at (1, \tan\theta), at distance |\sec\theta| from the origin, and meets y = 1 at (\cot\theta, 1), at distance |\csc\theta|. A positive secant means the beam itself reaches x = 1; a negative one means only its backward extension does.

Why it is true

The gaps. Each of tangent and secant divides by x, and x = 0 exactly when the terminal point is (0, 1) or (0, -1), at 90^\circ, 270^\circ and every angle a whole number of half turns from them. Cotangent and cosecant divide by y, which is 0 only at (1, 0) and (-1, 0), at the multiples of 180^\circ. Sine and cosine divide by nothing.

Secant and cosecant ranges. The terminal point is on the unit circle, so -1 \le x \le 1. Where the secant exists, x \ne 0, so 0 < |x| \le 1 and therefore \displaystyle |\sec\theta| = \frac{1}{|x|} \ge 1. Every such value is reached: given v with |v| \ge 1, the number \displaystyle \frac{1}{v} lies in [-1, 1], the vertical line \displaystyle x = \frac{1}{v} meets the circle, and the angle of that point has \sec\theta = v. The cosecant argument is the same with y.

Tangent and cotangent ranges. Given any number m, the line y = mx through the origin meets the unit circle at a point with x \ne 0, and the angle of that point has \tan\theta = m. For the cotangent use the line x = my, which meets the circle at a point with y \ne 0.

Distances. The terminal side lies on y = (\tan\theta)x, which at x = 1 has height \tan\theta. Dividing x^2 + y^2 = 1 by x^2 gives 1 + \tan^2\theta = \sec^2\theta, so the distance from the origin to (1, \tan\theta) is \displaystyle \sqrt{1 + \tan^2\theta} = |\sec\theta|. The point (1, \tan\theta) is the terminal point multiplied by \displaystyle \frac{1}{x} = \sec\theta; a positive multiplier keeps it on the beam and a negative one sends it to the opposite ray. Dividing by y^2 instead gives 1 + \cot^2\theta = \csc^2\theta and the same facts for the wall y = 1.

Worked examples

Example 1

All six from one point

The terminal point of \theta is \displaystyle \left(-\frac{65}{97}, \frac{72}{97}\right). Check that it is on the unit circle and find all six trigonometric values of \theta.

Solution

The point is on the circle because 65^2 + 72^2 = 4225 + 5184 = 9409 = 97^2, so the squares of the coordinates add to 1. Sine and cosine are read straight off: \displaystyle \sin\theta = \frac{72}{97} and \displaystyle \cos\theta = -\frac{65}{97}. Neither coordinate is 0, so all four quotients exist. Dividing fractions with the same denominator, the 97s cancel:

\begin{aligned} \tan\theta &= \frac{y}{x} = -\frac{72}{65} \\ \cot\theta &= \frac{x}{y} = -\frac{65}{72} \\ \sec\theta &= \frac{1}{x} = -\frac{97}{65} \\ \csc\theta &= \frac{1}{y} = \frac{97}{72} \end{aligned}

Each reciprocal keeps the sign of its partner. The secant and cosecant are both larger than 1 in size, as the range below says they must be.

Example 2

An angle on an axis

Decide which of the six functions are defined at \displaystyle \theta = \frac{7\pi}{2}, and give the values of those that are.

Solution

Subtracting one full turn, \displaystyle \frac{7\pi}{2} - 2\pi = \frac{3\pi}{2}, so the angle ends pointing straight down and its terminal point is (0, -1). Here x = 0, so the two functions that divide by x, \tan\theta and \sec\theta, are undefined. The other four are

\begin{aligned} \sin\theta &= -1 & \csc\theta &= \frac{1}{-1} = -1 \\ \cos\theta &= 0 & \cot\theta &= \frac{0}{-1} = 0 \end{aligned}

The cotangent is 0, not undefined: a 0 on top of a fraction is harmless, and only a 0 underneath is fatal. Notice also that \displaystyle \cot\theta = \frac{1}{\tan\theta} cannot be used here, since \tan\theta does not exist; \displaystyle \cot\theta = \frac{x}{y} still works.

Example 3

Listing the gaps

Find every \theta with -2\pi \le \theta \le 2\pi at which \csc\theta is undefined, and state the domain of the cosecant.

Solution

The cosecant is \displaystyle \frac{1}{y}, so it fails exactly when the terminal point is on the x-axis, at (1, 0) or (-1, 0). Those are the angles 0, \pi, 2\pi and so on in both directions: the integer multiples of \pi. Between -2\pi and 2\pi that gives five angles,

\theta = -2\pi,\ -\pi,\ 0,\ \pi,\ 2\pi

The domain of the cosecant is every real \theta except these and the other multiples: \theta \ne k\pi for every integer k.

Example 4

Tipene’s searchlight

Tipene’s searchlight stands 12 meters from a long straight wall. He aims the beam 35^\circ away from the perpendicular to the wall. How far along the wall from the foot of the perpendicular does the spot land, and how long is the beam, to the nearest centimeter?

Solution

Put the light at the origin with the wall along x = 12. This is the picture of Figure 1 scaled up by 12, so the spot is at height 12\tan 35^\circ and the beam length is 12\sec 35^\circ. A calculator finds the secant as one over the cosine:

\begin{aligned} 12\tan 35^\circ &\approx 12(0.70021) \approx 8.40 \\ 12\sec 35^\circ &= \frac{12}{\cos 35^\circ} \approx \frac{12}{0.81915} \approx 14.65 \end{aligned}

The spot is about 8.40 meters along the wall and the beam is about 14.65 meters long. As a check, 12^2 + 8.40^2 \approx 214.6 and 14.65^2 \approx 214.6.

Example 5

Values that cannot happen

Is there an angle with \sec\theta = 0.4? Find every angle with 0^\circ \le \theta < 360^\circ and \csc\theta = -1.25, to the nearest hundredth of a degree.

Solution

A secant of 0.4 would need \displaystyle \cos\theta = \frac{1}{0.4} = 2.5, but every point of the unit circle has |x| \le 1. No angle has \sec\theta = 0.4; the value lies in the gap of the range.

The value -1.25 is in the range. It means \displaystyle \sin\theta = \frac{1}{-1.25} = -0.8, and the horizontal line y = -0.8 meets the unit circle twice, once in Quadrant III and once in Quadrant IV. The acute angle with sine 0.8 is about 53.13^\circ, so the two angles are

\begin{aligned} \theta &\approx 180^\circ + 53.13^\circ = 233.13^\circ \\ \theta &\approx 360^\circ - 53.13^\circ = 306.87^\circ \end{aligned}

Example 6

Five values from one

\displaystyle \csc\theta = -\frac{29}{21} and \cot\theta > 0. Find the other five values.

Solution

The sine is the reciprocal, \displaystyle \sin\theta = -\frac{21}{29}. The terminal point is on the unit circle, so

\cos^2\theta = 1 - \frac{441}{841} = \frac{400}{841}, \qquad \cos\theta = \pm\frac{20}{29}

The cotangent is \displaystyle \frac{\cos\theta}{\sin\theta} and is positive, so the cosine has the same sign as the sine, which is negative: \displaystyle \cos\theta = -\frac{20}{29}, a terminal point in Quadrant III (signs by quadrant are in Signs of the trig functions by quadrant). Then

\begin{gathered} \tan\theta = \frac{21}{20} \qquad \sec\theta = -\frac{29}{20} \\ \cot\theta = \frac{20}{21} \end{gathered}

Example 7

A negative secant on the wall

Whetu’s searchlight is 10 meters from a straight wall along x = 10. She turns it to \theta = 2.5 radians. Where does the line of the beam meet the wall, how far is that point from the light, and what does the sign of \sec 2.5 say?

Solution

Since \displaystyle \frac{\pi}{2} \approx 1.571 < 2.5 < \pi, the beam points up and to the left, away from the wall. The line it lies on is y = (\tan 2.5)x, and at x = 10 it has height

10\tan 2.5 \approx 10(-0.74702) \approx -7.47

The distance from the light to (10, -7.47) is \displaystyle 10\sqrt{1 + \tan^2 2.5} = 10\,|\sec 2.5|, and in radian mode

\begin{aligned} \sec 2.5 &= \frac{1}{\cos 2.5} \approx -1.248 \\ 10\,|\sec 2.5| &\approx 12.48 \end{aligned}

The meeting point is about 7.47 meters below the foot of the perpendicular, 12.48 meters from the light. The negative secant says the beam never gets there: (10, -7.47) is 10\sec 2.5 times the terminal point, and a negative multiple lies on the opposite ray.

Boundaries

Undefined is not zero. At 90^\circ the cotangent is 0 and the tangent has no value.

\displaystyle \cot\theta = \frac{1}{\tan\theta} has holes. It fails at the odd multiples of 90^\circ, where the cotangent is 0 but the tangent does not exist. The form \displaystyle \cot\theta = \frac{\cos\theta}{\sin\theta} holds everywhere the cotangent is defined.

The ranges are for the plain functions. A stretched or shifted function such as 3\sec\theta has a different range, (-\infty, -3] \cup [3, \infty); graphs of such functions are in Graphing tangent and the reciprocal functions.

Radians and degrees. The domain of tangent is \displaystyle \theta \ne \frac{\pi}{2} + k\pi in radians and \theta \ne 90^\circ + k \cdot 180^\circ in degrees; \tan 90, in radians, exists.

Where this goes next

The undefined angles become the vertical asymptotes of the graphs of tangent, secant, cosecant and cotangent in Graphing tangent and the reciprocal functions, and the ranges become the gaps those graphs never enter. The reciprocal and quotient forms, domains attached, begin the Reciprocal and quotient identities, and 1 + \tan^2\theta = \sec^2\theta grows out of The Pythagorean identities. Exact values at the axis and special angles are in Exact values on the unit circle, and every value repeats after a full turn (periodicity).

Exercises

Six values from a point

Find all six trigonometric values of \theta.

The terminal point of \theta is \displaystyle \left(-\frac{99}{101}, -\frac{20}{101}\right).
The terminal point of \theta is \displaystyle \left(\frac{60}{109}, -\frac{91}{109}\right).
The terminal side of \theta passes through (-28, 45).
The terminal side of \theta passes through (-1, 2). Give exact values with rational denominators.

Where each function is defined

Which of the six functions are undefined at \theta = 5\pi? Give the values of the others.
Which of the six functions are undefined at \theta = -630^\circ? Give the values of the others.
List every \theta with 0^\circ \le \theta < 720^\circ at which \sec\theta is undefined.
List every \theta with -\pi \le \theta \le 3\pi at which \cot\theta is undefined.
State the domains of \tan\theta and of \csc\theta, in radians, and explain each.

Ranges

Which of these can happen for some angle? (a) \sec\theta = -0.5; (b) \csc\theta = 3; (c) \cot\theta = -1000; (d) \sec\theta = 1; (e) \csc\theta = 0.
Find every \theta with 0^\circ \le \theta < 360^\circ and \csc\theta = 2.5, to the nearest hundredth of a degree.
Find every \theta with 0^\circ \le \theta < 360^\circ and \sec\theta = -1.6, to the nearest hundredth of a degree.

One value given

\displaystyle \sec\theta = \frac{85}{36} and \sin\theta < 0. Find the other five values.
\displaystyle \cot\theta = \frac{16}{63} and \csc\theta < 0. Find the other five values.

The searchlight

Each light shines on a long straight wall; angles are measured from the perpendicular to the wall. Give lengths to the nearest centimeter.

Whetu’s light is 15 meters from the wall and aimed at 50^\circ. How far along the wall is the spot, and how long is the beam?
Eilis’s light is 8 meters from the wall and the beam is 20 meters long. Find the angle to the nearest hundredth of a degree and how far along the wall the spot is.
Eilis moves her light to 10 meters from the wall. How far along the wall is the spot at 89^\circ and at 89.9^\circ? What happens at 90^\circ, and which two functions does that make undefined?
Tipene’s light is 12 meters from the wall along x = 12 and is turned to 200^\circ. Where does the line of the beam meet the wall, how far is that from the light, and does the beam itself get there?

Reasoning

Explain why \tan\theta and \sec\theta are undefined at exactly the same angles, and why \cot\theta and \csc\theta are too.
Explain why no angle has |\csc\theta| < 1.
Explain why every real number is a value of \cot\theta.
Tipene says \displaystyle \cot\theta = \frac{1}{\tan\theta} for every angle where \cot\theta is defined. Give an angle where he is wrong, and a form of the cotangent that has no such exception.
Explain why \sec\theta = 1 only for angles coterminal with 0^\circ, and what the searchlight is doing then.
Show that 1 + \cot^2\theta = \csc^2\theta for every \theta at which both sides are defined.
Explain why \sec\theta and \csc\theta are never 0, while \tan\theta and \cot\theta are 0 at some angles.

Answers

\displaystyle \sin\theta = -\frac{20}{101}, \displaystyle \cos\theta = -\frac{99}{101}, \displaystyle \tan\theta = \frac{20}{99}, \displaystyle \cot\theta = \frac{99}{20}, \displaystyle \sec\theta = -\frac{101}{99}, \displaystyle \csc\theta = -\frac{101}{20}. The point checks: 99^2 + 20^2 = 10{,}201 = 101^2.
\displaystyle \sin\theta = -\frac{91}{109}, \displaystyle \cos\theta = \frac{60}{109}, \displaystyle \tan\theta = -\frac{91}{60}, \displaystyle \cot\theta = -\frac{60}{91}, \displaystyle \sec\theta = \frac{109}{60}, \displaystyle \csc\theta = -\frac{109}{91}.
Here \displaystyle r = \sqrt{784 + 2025} = 53. \displaystyle \sin\theta = \frac{45}{53}, \displaystyle \cos\theta = -\frac{28}{53}, \displaystyle \tan\theta = -\frac{45}{28}, \displaystyle \cot\theta = -\frac{28}{45}, \displaystyle \sec\theta = -\frac{53}{28}, \displaystyle \csc\theta = \frac{53}{45}.
Here \displaystyle r = \sqrt{5}. \displaystyle \sin\theta = \frac{2\sqrt{5}}{5}, \displaystyle \cos\theta = -\frac{\sqrt{5}}{5}, \tan\theta = -2, \displaystyle \cot\theta = -\frac{1}{2}, \displaystyle \sec\theta = -\sqrt{5}, \displaystyle \csc\theta = \frac{\sqrt{5}}{2}.
\csc and \cot are undefined. 5\pi - 4\pi = \pi, so the terminal point is (-1, 0) and y = 0. The others: \sin 5\pi = 0, \cos 5\pi = -1, \tan 5\pi = 0, \sec 5\pi = -1.
\tan and \sec are undefined. -630^\circ + 720^\circ = 90^\circ, so the terminal point is (0, 1) and x = 0. The others: \sin = 1, \cos = 0, \csc = 1, \cot = 0.
90^\circ, 270^\circ, 450^\circ, 630^\circ: the angles 90^\circ + k \cdot 180^\circ in the interval, where x = 0.
-\pi, 0, \pi, 2\pi, 3\pi: the multiples of \pi in the interval, where y = 0.
Tangent: \displaystyle \theta \ne \frac{\pi}{2} + k\pi, since it divides by x, which is 0 only at (0, \pm 1). Cosecant: \theta \ne k\pi, since it divides by y, which is 0 only at (\pm 1, 0). In both, k is any integer.
(a) No: it needs \cos\theta = -2. (b) Yes: \displaystyle \sin\theta = \frac{1}{3}. (c) Yes: cotangent takes every real value. (d) Yes: at \theta = 0^\circ. (e) No: \displaystyle \frac{1}{y} is never 0.
23.58^\circ and 156.42^\circ. \displaystyle \sin\theta = \frac{1}{2.5} = 0.4, the acute angle is about 23.578^\circ, and the Quadrant II angle is 180^\circ - 23.578^\circ.
128.68^\circ and 231.32^\circ. \displaystyle \cos\theta = \frac{1}{-1.6} = -0.625, and the calculator gives about 128.682^\circ; the other is 360^\circ - 128.682^\circ.
\displaystyle \cos\theta = \frac{36}{85}, \displaystyle \sin\theta = -\frac{77}{85}, \displaystyle \tan\theta = -\frac{77}{36}, \displaystyle \csc\theta = -\frac{85}{77}, \displaystyle \cot\theta = -\frac{36}{77}. Here \displaystyle \sin^2\theta = 1 - \frac{1296}{7225} = \frac{5929}{7225}, and the sine is negative.
\displaystyle \sin\theta = -\frac{63}{65}, \displaystyle \cos\theta = -\frac{16}{65}, \displaystyle \tan\theta = \frac{63}{16}, \displaystyle \sec\theta = -\frac{65}{16}, \displaystyle \csc\theta = -\frac{65}{63}. A negative cosecant makes y < 0, and a positive cotangent then makes x < 0: the point (-16, -63) is on the terminal side, with r = 65.
Spot about 17.88 meters along, beam about 23.34 meters. 15\tan 50^\circ \approx 17.876 and \displaystyle \frac{15}{\cos 50^\circ} \approx 23.336.
\theta \approx 66.42^\circ, spot about 18.33 meters along. \displaystyle \sec\theta = \frac{20}{8} = 2.5, so \cos\theta = 0.4; the spot is \displaystyle \sqrt{20^2 - 8^2} = \sqrt{336} \approx 18.330.
About 572.90 meters at 89^\circ and 5729.57 meters at 89.9^\circ (10\tan 89^\circ \approx 572.900, 10\tan 89.9^\circ \approx 5729.572). At 90^\circ the beam is parallel to the wall and never meets it: \tan 90^\circ and \sec 90^\circ are undefined.
At (12, 4.37), about 12.77 meters from the light; the beam itself does not get there. 12\tan 200^\circ \approx 4.368 and \sec 200^\circ \approx -1.0642, so the distance is 12\,|\sec 200^\circ| \approx 12.770. The negative secant puts the point on the opposite ray, at 20^\circ.
Both \displaystyle \tan\theta = \frac{y}{x} and \displaystyle \sec\theta = \frac{1}{x} divide by x and by nothing else, so each fails exactly when x = 0. Likewise \displaystyle \cot\theta = \frac{x}{y} and \displaystyle \csc\theta = \frac{1}{y} fail exactly when y = 0.
y is a coordinate of a point on the unit circle, so |y| \le 1, and where the cosecant exists y \ne 0. Then \displaystyle |\csc\theta| = \frac{1}{|y|} \ge 1.
Given any number c, the line x = cy passes through the origin and meets the unit circle at a point with y \ne 0 (setting y = 0 would force x = 0, which is not on the circle). The angle of that point has \displaystyle \cot\theta = \frac{x}{y} = c.
At \theta = 90^\circ, \displaystyle \cot 90^\circ = \frac{0}{1} = 0 but \tan 90^\circ does not exist, so \displaystyle \frac{1}{\tan\theta} has no value. The form \displaystyle \cot\theta = \frac{\cos\theta}{\sin\theta} works at every angle where the cotangent is defined.
\sec\theta = 1 means x = 1, and the only point of the unit circle with x = 1 is (1, 0), the terminal point of 0^\circ and its coterminal angles. The beam then points straight at the wall, along the perpendicular, and has its shortest possible length, 1.
Where y \ne 0, divide x^2 + y^2 = 1 by y^2: \displaystyle \frac{x^2}{y^2} + 1 = \frac{1}{y^2}, which is \cot^2\theta + 1 = \csc^2\theta.
A fraction is 0 only when its numerator is 0. The numerators of \displaystyle \frac{1}{x} and \displaystyle \frac{1}{y} are 1, so the secant and cosecant are never 0. The numerators of \displaystyle \frac{y}{x} and \displaystyle \frac{x}{y} are coordinates, which can be 0: \tan 0^\circ = 0 and \cot 90^\circ = 0.