Quadrantal angles: where tan, cot, sec and csc break

Explain it like I am five

Picture a clock hand that starts at 3 o'clock, pointing right; positive angles turn it counterclockwise, negative ones clockwise. A quadrantal angle stops the hand on an axis, one of the two lines through the center, at one of four points of the unit circle: (1, 0) at 3 o'clock, (0, 1) at 12, (−1, 0) at 9, (0, −1) at 6. Each point lists x first, then y.

Example: 810°. Remove two full laps: 810° − 720° = 90°, so the hand stops at 12 o'clock, where x = 0 and y = 1. Tangent, tan = $\frac{y}{x}$ , and secant, sec = $1 over x$ , both give $1 over 0$ : no answer. Cotangent, cot = $\frac{x}{y}$ = $0 over 1$ = 0, and cosecant, csc = $1 over y$ = 1, are fine.

Why no answer? $1 over 0$ asks what number times 0 gives 1, and every number times 0 is 0. That is undefined. A 0 on top is fine. So tan and sec break where x = 0, at 12 and 6 o'clock; cot and csc break where y = 0, at 3 and 9.

In plain words

Picture the rider stopping exactly above, below, right of, or left of the wheel's center. These four positions lie on the coordinate axes, the two lines through the center. A quadrantal angle ends at one of those positions. Four trigonometric functions use division: tangent uses height divided by horizontal position, cotangent uses the opposite ratio, secant uses 1 divided by horizontal position, and cosecant uses 1 divided by height. Some of these divisions fail at an axis because the bottom becomes 0. The word undefined means there is no output for that input. The numerator, the number on top, may be 0 without causing that failure.

The graph of tan t breaks at the dashed lines t = ± and t = ±, exactly where x = cos t is 0.

Reminder

Coterminal reduction. A full turn leaves the terminal point unchanged: 630° − 360° = 270°.
Zero numerator. $0 over −1$ = 0 because 0 × (−1) = 0; $1 over 0$ has no value.

The graph of tan t breaks at the dashed lines t = ± and t = ±, exactly where x = cos t is 0.

This reference row shows tan at the four axis positions.

This reference row shows cot at the four axis positions.

This reference row shows sec at the four axis positions.

This reference row shows csc at the four axis positions.

Read the output entry in the column under its input or category.

Why it works. Dividing 1 by 0 would require a number q with q × 0 = 1, but every number times 0 equals 0. At a unit-circle axis point, the nonzero coordinate is 1 or −1, so any zero denominator has a nonzero numerator. Horizontal position x is 0 at the top and bottom, a half turn apart. Height y is 0 at the right and left, also a half turn apart. These facts locate every excluded input, including negative angles and angles after many turns.

Ruletan t and sec t: t ≠ + nπ, or t ≠ 90° + n·180°. cot t and csc t: t ≠ nπ, or t ≠ n·180°. An integer is a number such as −2, −1, 0, 1 or 2; n may be any integer. The notes write the same exclusions as ± nπ and ±nπ. Every other real input is allowed.

The same idea, five ways

Say it

An excluded input makes the bottom of the function's fraction zero.

Write it

Tangent and secant exclude the y-axis angles; cotangent and cosecant exclude the x-axis angles.

In math

tan t = $\frac{y}{x}$ and sec t = $1 over x$ require x ≠ 0.
cot t = $\frac{x}{y}$ and csc t = $1 over y$ require y ≠ 0.
tan and sec: t ≠ $π over 2$ + nπ.
cot and csc: t ≠ nπ.
On the graph, an excluded input has no point.
{t | t ≠ $π over 2$ + nπ for every integer n}: tan and sec inputs.
{t | t ≠ nπ for every integer n}: cot and csc inputs.

An input that asks for division by zero is like asking to share one item among zero groups.

See it

At this y-axis point, x = 0 removes tangent and secant from their domains.

The same idea, other ways

As failed sharing

Dividing 1 item among 0 groups cannot produce a number per group: multiplying any proposed answer by 0 still gives 0 items.

There is no group size that makes zero groups contain one item.

From the wheel

At the top and bottom, horizontal position is 0, so any formula dividing by x fails. At the right and left, height is 0, so any formula dividing by y fails.

At the top, x = 0 and y = 1, so tangent and secant have a zero denominator.

t	P(x, y)	tan t = $\frac{y}{x}$	cot t = $\frac{x}{y}$	sec t = $1 over x$	csc t = $1 over y$
0 (0°)	(1, 0)	0	undefined	1	undefined
$π over 2$ (90°)	(0, 1)	undefined	0	undefined	1
π (180°)	(−1, 0)	0	undefined	−1	undefined
$3 π over 2$ (270°)	(0, −1)	undefined	0	undefined	−1

.1Tangent divides by horizontal position

Tangent compares height with horizontal reach. The denominator is x, so its failure points are on the vertical axis.

tan t = $\frac{y}{x}$ = $\frac{\sin t}{\cos t}$ .
Domain excludes $π over 2$ + nπ, with integer n.

Horizontal position is 0 at the vertical-axis positions.

Worked exampleA defined zero tangent

The question asks whether tangent exists at 720°, and what its value is.

The height is 0, but the horizontal denominator is 1.

720° − 2 × 360° = 0°.Remove two full turns to locate the same terminal point.
The point is (1, 0), so tan 720° = $0 over 1$ = 0.The denominator is 1, so dividing is allowed.

Answer

tan 720° = 0.

Check The value 0 times the denominator 1 reproduces the numerator 0.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: A 0 on top makes this function undefined.

A zero numerator is allowed if the bottom is nonzero.

✓ Instead: Write the definition and check its denominator; = 0.

Tips and tricks

Use this object's defining formula to check the example before relying on a remembered pattern.

.2Cotangent divides by height

Cotangent compares horizontal reach with height. Its denominator is y, so it fails on the horizontal axis.

cot t = $\frac{x}{y}$ = $\frac{\cos t}{\sin t}$ .
Domain excludes nπ, with integer n.
cot t = $1 over tan t$ applies where both functions are defined and tan t ≠ 0. The coordinate ratio still defines cot t at y-axis points where tan t fails.

Height is 0 at the horizontal-axis positions.

Worked exampleCotangent can be zero

The question asks whether cotangent exists at 630°, and what its value is.

The cotangent numerator is 0 while its height denominator is −1.

630° − 360° = 270°.One full turn gives the same point inside a single turn.
The point is (0, −1), so cot 630° = $0 over −1$ = 0.The denominator is −1, so cotangent is defined.

Answer

cot 630° = 0.

Check At the same angle, tangent is and is undefined. Use cotangent's own ratio, rather than taking a reciprocal of an undefined value.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: A 0 on top makes this function undefined.

A zero numerator is allowed if the bottom is nonzero.

✓ Instead: Write the definition and check its denominator; = 0.

Tips and tricks

Use this object's defining formula to check the example before relying on a remembered pattern.

.3Secant divides 1 by horizontal position

Secant flips the horizontal coordinate. It fails wherever that coordinate is 0, exactly where tangent fails.

sec t = $1 over x$ = $1 over cos t$ .
Domain excludes $π over 2$ + nπ, with integer n.

Here x = 0, so secant cannot divide 1 by x.

Worked exampleSecant at a vertical-axis point

The question asks whether secant exists at 450°.

The horizontal coordinate 0 makes secant undefined.

450° − 360° = 90°.Remove one full turn to identify the terminal point.
At (0, 1), sec 450° would be $1 over 0$ .Secant divides 1 by the horizontal coordinate x.
Report undefined.No number multiplied by 0 gives the numerator 1.

Answer

sec 450° is undefined.

Check 450° = 90° + 2 × 180°, so it belongs to secant's excluded input list.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: A 0 on top makes this function undefined.

A zero numerator is allowed if the bottom is nonzero.

✓ Instead: Write the definition and check its denominator; = 0.

Tips and tricks

Use this object's defining formula to check the example before relying on a remembered pattern.

.4Cosecant divides 1 by height

Cosecant flips the height coordinate. It fails wherever the height is 0, exactly where cotangent fails.

csc t = $1 over y$ = $1 over sin t$ .
Domain excludes nπ, with integer n.

Here y = 0, so cosecant cannot divide 1 by y.

Worked exampleCosecant at a horizontal-axis point

The question asks whether cosecant exists at 720°.

The height denominator is 0.

720° − 2 × 360° = 0°.Removing two turns leaves the same terminal point.
At (1, 0), csc 720° would be $1 over 0$ .Cosecant divides 1 by height y.
Report undefined.No number times 0 gives 1.

Answer

csc 720° is undefined.

Check 720° is a multiple of 180°, so it is on cosecant's excluded input list.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: A 0 on top makes this function undefined.

A zero numerator is allowed if the bottom is nonzero.

✓ Instead: Write the definition and check its denominator; = 0.

Tips and tricks

Use this object's defining formula to check the example before relying on a remembered pattern.

Strategy: step by step

1. If the angle is negative or at least a full turn, add or subtract full turns (2π or 360°) to bring it between 0 and 2π.
2. Check whether it is a quarter-turn mark: 0 gives (1, 0), $π over 2$ (90°) gives (0, 1), π (180°) gives (−1, 0), $3 π over 2$ (270°) gives (0, −1). If it is none of these, all six functions are defined.
3. On the y-axis (x = 0): tan and sec are undefined, while cot and csc have values.
4. On the x-axis (y = 0): cot and csc are undefined, while tan and sec have values. In a table, write 'undefined' or 'n/a', as the notes do.
Read a reference-table entry in the column under its input or category in the matching picture.

Strategy

Finding whether a trigonometric function is defined

Does the terminal side lie on an axis?

YesRead the point from (1, 0), (0, 1), (−1, 0), (0, −1).

NoBoth x and y are nonzero, so all six functions are defined.

↓

Is the requested fraction's denominator 0?

YesExclude that input from the domain and write undefined.

NoThe function is defined. A zero numerator gives 0.

1. Reduce the angle by full turns to the interval [0, 2π), or [0°, 360°).
2. Find its terminal point if it is an axis angle.
3. Write the requested function as a fraction of x and y.
4. If the denominator is 0, report undefined. Otherwise divide.

Worked exampleWhich functions are undefined?

The question asks which division formulas have a zero denominator at each angle. Which of tan t, cot t, sec t and csc t are undefined at (a) t = 3π and (b) t = −90°?

At 3π, the terminal point is (−1, 0), so formulas dividing by y fail.

At −90°, the terminal point is (0, −1), so formulas dividing by x fail.

(a) 3π is more than 2π, so subtract one full turn: 3π − 2π = π.A full turn lands on the same point, so 3π and π share their point.
The point for π is (−1, 0): x = −1 and y = 0.Half a turn from (1, 0) reaches the far left of the circle.
y = 0 is on the bottom of cot t = $\frac{x}{y}$ and csc t = $1 over y$ , so those two are undefined. The other two have values: tan 3π = $0 over −1$ = 0 and sec 3π = $1 over −1$ = −1.Only a 0 on the bottom breaks a fraction; a 0 on top gives the value 0.
(b) −90° is a quarter turn clockwise, ending at the bottom of the circle, (0, −1): x = 0 and y = −1.Negative angles turn clockwise, and a quarter turn is 90°.
x = 0 is on the bottom of tan t = $\frac{y}{x}$ and sec t = $1 over x$ , so those two are undefined. cot(−90°) = $0 over −1$ = 0 and csc(−90°) = $1 over −1$ = −1.Same rule: a 0 on the bottom means undefined.

Answer

(a) cot 3π and csc 3π are undefined.
(b) tan(−90°) and sec(−90°) are undefined.

Check Use the excluded lists: 3π = nπ with n = 3, which is excluded for cot and csc; −90° = 90° + n·180° with n = −1, which is excluded for tan and sec.

Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.

Rung 1Which of tan, cot, sec and csc break at − and at 540°?

For each angle, decide which of tan, cot, sec and csc are defined. Give the value of each one that is defined. Write 'undefined' for each one that is not. (a) t = − $5 π over 2$ (b) t = 540°

(a) Add full turns to − $5 π over 2$ . First get − $5 π over 2$ + 2π = − $π over 2$ . Then get − $π over 2$ + 2π = $3 π over 2$ .Adding 2π gives the same point on the unit circle. Two full turns bring the angle between 0 and 2π.
(a) Note that $3 π over 2$ is a quarter-turn mark. Its point is (0, −1), so x = 0 and y = −1. $3 π over 2$ (270°) lands on the negative y-axis.
(a) Take tan t = $\frac{y}{x}$ = $−1 over 0$ and sec t = $1 over x$ = $1 over 0$ . Both are undefined.On the y-axis x = 0, and x is the denominator of tan and sec.
(a) Take cot t = $\frac{x}{y}$ = $0 over −1$ = 0 and csc t = $1 over y$ = $1 over −1$ = −1.For cot and csc the denominator is y = −1, which is not 0. So both have values.
(b) Subtract one full turn: 540° − 360° = 180°.The angle is at least a full turn, so subtract 360° to bring it between 0° and 360°.
(b) Note that 180° is a quarter-turn mark. Its point is (−1, 0), so x = −1 and y = 0.180° (π) lands on the negative x-axis.
(b) Take cot t = $\frac{x}{y}$ = $−1 over 0$ and csc t = $1 over y$ = $1 over 0$ . Both are undefined.On the x-axis y = 0, and y is the denominator of cot and csc.
(b) Take tan t = $\frac{y}{x}$ = $0 over −1$ = 0 and sec t = $1 over x$ = $1 over −1$ = −1.For tan and sec the denominator is x = −1, which is not 0. So both have values.

Answer

(a) At t = − $5 π over 2$ : tan is undefined and sec is undefined
cot = 0 and csc = −1. (b) At t = 540°: cot is undefined and csc is undefined
tan = 0 and sec = −1.

Check Use the rule. (a) − = + nπ with n = −3, because − 3π = −. That is the exclusion for tan and sec. It is not of the form nπ, because − ÷ π = −2.5 is not an integer. So cot and csc are allowed. (b) 540° = n·180° with n = 3. That is the exclusion for cot and csc. Also (540° − 90°) ÷ 180° = 2.5 is not an integer, so tan and sec are allowed. Both parts agree with the answer.

Work to write

(a) − $5 π over 2$ + 2π + 2π = $3 π over 2$ , point (0, −1)
(a) x = 0 ⇒ tan = undefined, sec = undefined
(a) cot = $0 over −1$ = 0, csc = $1 over −1$ = −1
(b) 540° − 360° = 180°, point (−1, 0)
(b) y = 0 ⇒ cot = undefined, csc = undefined
(b) tan = $0 over −1$ = 0, sec = $1 over −1$ = −1

(a) At t = − $5 π over 2$ : tan is undefined and sec is undefined; cot = 0 and csc = −1. (b) At t = 540°: cot is undefined and csc is undefined; tan = 0 and sec = −1.

Rung 2Which of tan, cot, sec and csc break at t = ?

Let t = $11 π over 2$ . Find tan t, cot t, sec t and csc t. Write 'undefined' for any function with no value at this input.

The angle ends at the same place as 270°, at P(0, −1) on the negative y-axis. Here x = 0, so tan and sec are undefined.

Bring the angle into one turn: $11 π over 2$ − 2π − 2π = $11 π over 2$ − $8 π over 2$ = $3 π over 2$ . In degrees, 990° − 720° = 270°. $11 π over 2$ is at least a full turn. Adding or subtracting whole turns of 2π does not change the point on the unit circle.
Recognise $3 π over 2$ (270°) as a quarter-turn mark. Its point on the unit circle is (0, −1), so x = cos t = 0 and y = sin t = −1.The quarter-turn marks are 0 → (1, 0), $π over 2$ → (0, 1), π → (−1, 0) and $3 π over 2$ → (0, −1).
tan t = $\frac{y}{x}$ = $−1 over 0$ , which is undefined. sec t = $1 over x$ = $1 over 0$ , which is also undefined.The point lies on the y-axis, where x = 0. Tan and sec both divide by x, and a zero on the bottom is a break.
cot t = $\frac{x}{y}$ = $0 over −1$ = 0. csc t = $1 over y$ = $1 over −1$ = −1.Cot and csc divide by y = −1, which is not zero. A zero on top only makes the value 0.

Answer

tan is undefined, sec is undefined, cot = 0 and csc = −1.

Check Use the rule. Tan and sec exclude t = + nπ, and = + 5π with n = 5, so both are excluded. Cot and csc exclude t = nπ, but = 5.5π and 5.5 is not an integer, so both are defined. This agrees with the answer.

Work to write

$11 π over 2$ − 4π = $3 π over 2$ , which is coterminal with 270°
Point on the unit circle: (0, −1), so x = 0 and y = −1
tan t = $−1 over 0$ , so it is undefined
sec t = $1 over 0$ , so it is undefined
cot t = $0 over −1$ = 0
csc t = $1 over −1$ = −1

tan $11 π over 2$ is undefined, sec $11 π over 2$ is undefined, cot $11 π over 2$ = 0 and csc $11 π over 2$ = −1.

Rung 3Rung 3: reduce a degree angle

The question asks which of tangent and cosecant is undefined at 810°.

The top point excludes tangent but permits cosecant.

810° − 2 × 360° = 90°.Two full turns leave the same terminal point.
At (0, 1), tan = $1 over 0$ is undefined, while csc = $1 over 1$ = 1.Tangent divides by x = 0; cosecant divides by y = 1.

Answer

tan 810° is undefined.
csc 810° = 1.

Check 810° = 90° + 4 × 180° belongs to tangent's excluded input list.

Rung 4Solving for the inputs where tan(x + 30°) is undefined

Let f(x) = tan(x + 30°). Find every value of x with 0° ≤ x < 360° that is excluded from the domain of f, so that f(x) is undefined.

The angle u = x + 30° = 90° (when x = 60°) ends at P(0, 1) on the y-axis. There the x-coordinate is 0, so tan u = is undefined.

Call the angle inside the function u, so u = x + 30°. We need the values of u where tan u is undefined.The exclusion rule for tan applies to the whole angle tan acts on, not to x alone.
Write the exclusion: tan u is undefined exactly when u = 90° + n·180°, where n is any integer.tan u = $\frac{y}{x}$ uses the point (x, y) on the unit circle. It breaks on the y-axis, where x = 0. That happens at 90° and 270°, which are 180° apart.
Set x + 30° = 90° + n·180°. Subtract 30° from both sides to get x = 60° + n·180°.Solving the exclusion equation for x turns a statement about the angle into a statement about the input.
Try integer values of n. n = 0 gives x = 60°. n = 1 gives x = 240°. n = 2 gives x = 420°, which is too large. n = −1 gives x = −120°, which is too small.Only the x values with 0° ≤ x < 360° are asked for, so we keep the ones that land in that interval.
Check x = 60°: the angle is u = 60° + 30° = 90°, the quarter-turn mark at (0, 1). Check x = 240°: u = 240° + 30° = 270°, the quarter-turn mark at (0, −1).Both points lie on the y-axis, where x = 0. Step 3 of the method says tan is undefined there.

Answer

f(x) is undefined at x = 60° and x = 240°. In general, the excluded inputs are x = 60° + n·180°, where n is any integer.

Check At x = 60°, tan(90°) = , which is undefined. At x = 240°, tan(270°) = , which is also undefined. A nearby value has a value: x = 0° gives tan 30° = , so f is defined there. That is consistent with exclusions only at 60° and 240°.

Work to write

u = x + 30°
tan u is undefined when u = 90° + n·180°
x + 30° = 90° + n·180°
x = 60° + n·180°
n = 0: x = 60°; n = 1: x = 240°
x = 60° gives u = 90°, point (0, 1); x = 240° gives u = 270°, point (0, −1)
Excluded: x = 60° and x = 240°

f(x) is undefined at x = 60° and x = 240°. In general, the excluded inputs are x = 60° + n·180°, where n is any integer.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: tan π is undefined because its numerator is 0.

A zero numerator gives the output 0 if the denominator is nonzero.

✓ Instead: tan π = = 0; cot π has a zero denominator and is undefined.

Tips and tricks

Write the denominator before deciding whether the function exists: tan and sec divide by x; cot and csc divide by y.
Axes belong to neither neighboring quadrant. Check them separately.

Trap. Calling a 0 on top 'undefined'. = 0 is an ordinary value. Only a 0 on the bottom breaks a function, so tan π = 0 is defined while cot π is undefined.

Keep in mind

A handwritten one-page cheat sheet rebuilds every domain yes or no from three lines: the axis points 0° (1, 0), 90° (0, 1), 180° (−1, 0), 270° (0, −1), with π = 180°; sin = y, cos = x, tan = $\frac{y}{x}$ , cot = $\frac{x}{y}$ , sec = $1 over x$ , csc = $1 over y$ ; and 'zero on the bottom means undefined'.
Sine and cosine never break, because sin t = y and cos t = x involve no division: at 12 o'clock they are 1 and 0.
Remove laps before reading the point: −630° + 720° = 90°, so the hand points to 12 o'clock and tan(−630°) is undefined.
An axis angle sits in no quadrant, so read its point instead of the quadrant sign table: 0° is not in Quadrant I.

Memory hookNo 'co' (tan, sec): x on the bottom, broken at 12 and 6 o'clock. With 'co' (cot, csc): y on the bottom, broken at 3 and 9.

Flash cards: say the answer out loud, then flip

What is a quadrantal angle?

An angle that ends on an axis: a multiple of 90°, or of .

Why is undefined?

No number times 0 gives 1.

Let t = 7π. Find tan t, cot t, sec t and csc t. Write 'undefined' for any function with no value at this input.

7π − 6π = π, so the point is (−1, 0), which lies on the x-axis. tan t = 0, sec t = −1, and cot t and csc t are undefined.

Is −990° in the domain of sec?

No: −990° + 1080° = 90°, where x = 0, and sec = .

Find cot 1530° and csc 1530°.

1530° − 4 × 360° = 90°, the point (0, 1).
cot 1530° = $0 over 1$ = 0.
csc 1530° = $1 over 1$ = 1.

Is cot 810° undefined because 810° ends on an axis?

No: at (0, 1), cot = = 0. Only a 0 on the bottom breaks a function.