The range of tan, cot, sec and csc

Explain it like I am five

Picture a sign giving a hill's steepness: flat is 0, a gentle climb 0.1, a cliff 50, downhill negative. Tangent, tan t = $\frac{y}{x}$ , is the steepness of the line from the center to the point, so it can be any number; so can cotangent, $\frac{x}{y}$ .

Secant and cosecant are flips: sec t = $1 over cos t$ and csc t = $1 over sin t$ . A flip, or reciprocal, is 1 divided by the number. Flipping a number no bigger than 1 in size gives one at least 1 in size: 1 ÷ 0.5 = 2 and 1 ÷ 0.1 = 10.

Example: can sec t = 0.4? Flip it back: cos t would be 1 ÷ 0.4 = 2.5, bigger than 1, so no. Can csc t = −2.5? Flip: sin t = 1 ÷ (−2.5) = −0.4, an allowed sine, so yes.

So the range, every possible output, of tan and cot is (−∞, ∞): all real numbers (∞ is infinity). For sec and csc it is (−∞, −1] ∪ [1, ∞), read '−1 or less, or 1 or more'.

In plain words

Think of pouring one full measuring cup into smaller scoops. A half-cup scoop fits twice, and a tenth-cup scoop fits ten times. That is how secant and cosecant work: each divides 1 by a coordinate whose size is at most 1. Their outputs therefore have size at least 1. They can be positive or negative, but they never fall strictly between −1 and 1. Tangent and cotangent compare one coordinate with the other. There is no limit on that comparison, because one coordinate may be very small while the other stays large. Their range, the list of possible outputs, includes every real number.

The dashed cosine wave stays between −1 and 1; its reciprocal, sec t (solid), stays on or outside the lines y = 1 and y = −1.

Reminder

Reciprocal. 1 ÷ $1 over 2$ = 2 and 1 ÷ (− $1 over 2$ ) = −2; the reciprocal keeps the sign.
Union. (−∞, −1] ∪ [1, ∞) means an output at most −1 or at least 1.

The dashed cosine wave stays between −1 and 1; its reciprocal, sec t (solid), stays on or outside the lines y = 1 and y = −1.

The two closed endpoints belong to the secant and cosecant range, and the middle gap does not.

Nonzero coordinate values of size at most 1 flip to outputs of size at least 1.

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

Why it works. A nonzero cosine x with size at most 1 gives 1 ÷ x with size at least 1. Every output of size at least 1 occurs: its reciprocal is a nonzero number in [−1, 1], and some point has that cosine. The same argument works for cosecant and sine. For any proposed tangent output m, choose x = 1 ÷ and y = m ÷ . Then + = 1 and y ÷ x = m. Exchanging these coordinates proves every real cotangent output occurs too.

Ruletan t and cot t have range (−∞, ∞). sec t and csc t have range (−∞, −1] ∪ [1, ∞), meaning an output ≤ −1 or ≥ 1. To test a proposed secant or cosecant output v, first require v ≠ 0, then check that belongs to [−1, 1].

The same idea, five ways

Say it

Tangent and cotangent can give any real output; secant and cosecant have outputs of size one or larger.

Write it

The reciprocal coordinate functions never produce an output strictly between −1 and 1.

In math

tan and cot range: (−∞, ∞).
sec and csc range: (−∞, −1] ∪ [1, ∞).
For v = sec t or v = csc t: v ≤ −1 or v ≥ 1.
For v = sec t or v = csc t: |v| ≥ 1.
On their graphs, secant and cosecant have no point at an output height strictly between −1 and 1.
{v | v ≤ −1 or v ≥ 1}: sec and csc outputs.

A full cup contains at least one scoop whenever the scoop holds at most a full cup.

See it

The two closed endpoints belong to the secant and cosecant range, and the middle gap does not.

The same idea, other ways

With measuring scoops

One cup contains two half-cup scoops or ten tenth-cup scoops. Dividing 1 by a positive amount at most 1 produces an answer at least 1.

Smaller positive divisors produce larger reciprocal outputs.

With negative coordinates

If cosine is − $1 over 2$ , secant is −2. The sign stays negative while the size grows from one half to two.

A reciprocal keeps the sign and has size at least 1.

From the graph

Secant has a gap between output heights −1 and 1. Tangent crosses every possible output height on each unbroken graph piece between two neighboring excluded inputs.

Read the sec output vertically and the input horizontally.

Construct the output you want

For any real m, divide both (1, m) coordinates by r = $square root of 1 + m squared$ . The new squared coordinates sum to $1 over 1 + m squared$ + $m squared over 1 + m squared$ = 1, and their ratio is m. Swapping them makes cotangent equal m. Because 1 + $m^{2}$ > 0, the radius and the ratio's denominator are nonzero.

For m = 3, (1, 3) points in a direction whose tangent ratio is 3.

Function	Domain	Range
sin t	all real numbers	[−1, 1]
cos t	all real numbers	[−1, 1]
tan t	t ≠ $π over 2$ + nπ (t ≠ 90° + n·180°)	(−∞, ∞)
cot t	t ≠ nπ (t ≠ n·180°)	(−∞, ∞)
sec t	t ≠ $π over 2$ + nπ (t ≠ 90° + n·180°)	(−∞, −1] ∪ [1, ∞)
csc t	t ≠ nπ (t ≠ n·180°)	(−∞, −1] ∪ [1, ∞)

.1Tangent can produce every real output

You can choose coordinates with any desired ratio. One coordinate may be 0, provided the coordinate on the bottom is nonzero.

tan t = $\frac{y}{x}$ .
Range: (−∞, ∞).

The point (2, 5) gives ratio 5 divided by 2.

Worked exampleAn output for tan

The question asks whether 2.5 can be an output of tan t.

The point (2, 5) gives ratio 5 divided by 2.

For the proposed tangent output 2.5, use the terminal-side point (2, 5).This point has nonzero x, so tangent is defined.
Its ratio is $5 over 2$ = 2.5.Tangent is y divided by x; scaling the point onto the unit circle leaves the ratio unchanged.

Answer

tan t = 2.5 is possible.

Check After scaling (2, 5), the squared coordinates add to + = 1, so an actual unit-circle point has the requested tangent ratio.

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

✗ Not this: Every real proposed output is allowed for every trigonometric function.

Sine and cosine are bounded, and secant and cosecant exclude the middle gap.

✓ Instead: Use this function's stated range before accepting the output.

Tips and tricks

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

.2Cotangent can produce every real output

You can choose coordinates with any desired ratio. One coordinate may be 0, provided the coordinate on the bottom is nonzero.

cot t = $\frac{x}{y}$ .
Range: (−∞, ∞).

The point (−5, 2) gives cotangent ratio −5 divided by 2.

Worked exampleAn output for cot

The question asks whether −2.5 can be an output of cot t.

The point (−5, 2) gives cotangent ratio −5 divided by 2.

For the proposed cotangent output −2.5, use the terminal-side point (−5, 2).This point has nonzero y, so cotangent is defined.
Its ratio is $−5 over 2$ = −2.5.Cotangent is x divided by y; scaling the point keeps that ratio.

Answer

cot t = −2.5 is possible.

Check After scaling (−5, 2), the squared coordinates add to + = 1, so an actual unit-circle point has the requested cotangent ratio.

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

✗ Not this: Every real proposed output is allowed for every trigonometric function.

Sine and cosine are bounded, and secant and cosecant exclude the middle gap.

✓ Instead: Use this function's stated range before accepting the output.

Tips and tricks

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

.3Secant leaves a middle gap

The horizontal position has size at most 1. Its nonzero reciprocal therefore has size at least 1.

sec t = $1 over cos t$ .
Range: (−∞, −1] ∪ [1, ∞).

sec t =

cos t =

0 < < 1

The proposed secant flips to a possible cosine.

Worked exampleAn output for sec

The question asks whether $4 over 3$ can be an output of sec t.

sec t =

cos t =

0 < < 1

The proposed secant flips to a possible cosine.

If sec t = $4 over 3$ , then cos t = $3 over 4$ .Reciprocal partners flip one another wherever defined.
$3 over 4$ is nonzero and between −1 and 1.That is an allowed cosine, so the proposed secant occurs.

Answer

sec t = is possible.

Check The point (, ) is on U because + = 1. Its nonzero cosine gives secant .

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

✗ Not this: Every real proposed output is allowed for every trigonometric function.

Sine and cosine are bounded, and secant and cosecant exclude the middle gap.

✓ Instead: Use this function's stated range before accepting the output.

Tips and tricks

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

.4Cosecant leaves a middle gap

The height has size at most 1. Its nonzero reciprocal therefore has size at least 1.

csc t = $1 over sin t$ .
Range: (−∞, −1] ∪ [1, ∞).

csc t = −

sin t = −

−1 < − < 0

The proposed cosecant flips to a possible sine.

Worked exampleAn output for csc

The question asks whether − $4 over 3$ can be an output of csc t.

csc t = −

sin t = −

−1 < − < 0

The proposed cosecant flips to a possible sine.

If csc t = − $4 over 3$ , then sin t = − $3 over 4$ .Reciprocals keep their negative sign.
− $3 over 4$ is nonzero and lies between −1 and 1.That is an allowed sine, so the proposed cosecant occurs.

Answer

csc t = − is possible.

Check The point (, −) is on U because + = 1. Its nonzero sine − gives cosecant −.

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

✗ Not this: Every real proposed output is allowed for every trigonometric function.

Sine and cosine are bounded, and secant and cosecant exclude the middle gap.

✓ Instead: Use this function's stated range before accepting the output.

Tips and tricks

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

Strategy: step by step

1. For a tangent or cotangent range question, answer yes for every real output.
2. For secant or cosecant, first reject 0 because its reciprocal is undefined.
3. For any other proposed output, ignore its sign and check its distance from 0.
4. A size of at least 1 is allowed. A size less than 1 is impossible.
5. To verify the decision, flip a nonzero proposed output. It must become an allowed sine or cosine between −1 and 1.
Read a reference-table entry in the column under its input or category in the matching picture.

Strategy

Testing an output against a range

Is the function tangent or cotangent?

YesEvery real output is in its range.

NoFor secant or cosecant, continue to the size check.

↓

Is the proposed output 0?

YesReject it. A reciprocal of a nonzero coordinate never equals 0.

NoAccept if its distance from 0 is at least 1; otherwise reject.

1. Identify the requested function.
2. For tan or cot, accept every real proposed output.
3. For sec or csc, reject 0, then compare the output's size with 1.
4. As a second check for a nonzero output, take its reciprocal and compare with [−1, 1].

Worked exampleRange questions for the other four

The question asks whether each proposed output can occur. Answer yes or no. (a) Is 0.5 in the range of sec t? (b) Is −3 in the range of csc t? (c) Is 250 in the range of tan t? (d) Is −1 in the range of csc t?

The two closed endpoints belong to the secant and cosecant range, and the middle gap does not.

sec t = 0.5 would require cos t = 2

csc t = −3 requires sin t = −

csc t = −1 occurs at sin t = −1

Flipping a proposed output exposes whether its coordinate is possible.

(a) No. If sec t = 0.5, then cos t = 1 ÷ 0.5 = 2, and no point of the unit circle has x = 2.sec t and cos t are reciprocals, and cos t can never be bigger than 1.
(b) Yes. If csc t = −3, then sin t = − $1 over 3$ , which is between −1 and 1, so such a t exists.Flip the value back to a sine and check that it is an allowed sine.
(c) Yes. Tangent takes every real value.As the point approaches the top through Quadrant I, x approaches 0 through positive values while y approaches 1. The positive ratio y ÷ x therefore exceeds any fixed positive number; the point construction also supplies 250 exactly.
(d) Yes. csc t = −1 when sin t = −1, which happens at t = $3 π over 2$ (270°).−1 is an endpoint, and the bracket in (−∞, −1] includes it.

Answer

(a) no
(b) yes
(c) yes
(d) yes

Check For a nonzero secant or cosecant output, its reciprocal must be in [−1, 1]: 1 ÷ 0.5 = 2 fails; 1 ÷ (−3) = − passes; 1 ÷ (−1) = −1 passes. The proposed output 0 always fails before this test, because 1 ÷ 0 is undefined.

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

✗ Not this: sec t = 0 is possible because 0 is in the cosine range.

Secant is 1 divided by a nonzero cosine. Such a reciprocal cannot equal 0.

✓ Instead: 0 is outside the secant range. The flip test starts by rejecting 0.

Tips and tricks

Draw a gap between −1 and 1 for secant and cosecant; fill the endpoints.
Name the proposed output before testing it. Its range membership is separate from whether an angle is in the domain.

Trap. Writing the range of sec t or csc t as [−1, 1]. That is the range of their partners cos t and sin t. Flipping pushes every value outside: (−∞, −1] ∪ [1, ∞).

Keep in mind

Test a secant or cosecant value by its flip, not against [−1, 1]: sec t = 2.5 is possible because its flip, 0.4, is an allowed cosine, while sec t = 0.4 is not.
Only sine and cosine are stuck between −1 and 1: tan t = 40 and cot t = −40 are both possible.
Judge sec and csc by size, not sign: csc t = −1.5 is possible because its size, 1.5, is at least 1.
The edges count: csc t = −1 when sin t = −1, since 1 ÷ (−1) = −1, so the brackets at −1 and 1 are square.

Memory hookSlopes go anywhere; flips stay outside. Tan and cot take every number. To test a sec or csc value, flip it: the flip must land in [−1, 1].

Flash cards: say the answer out loud, then flip

What is a reciprocal?

1 divided by the number: the reciprocal of 0.05 is 20.

What does ∪ mean in (−∞, −1] ∪ [1, ∞)?

Union, read 'or': at most −1, or at least 1.

Range of tan t? Range of sec t?

tan t: (−∞, ∞).
sec t: (−∞, −1] ∪ [1, ∞).

Can csc t = 0.9?

No: the flip gives sin t = ≈ 1.11, bigger than 1.

Can sec t = −3?

Yes: the flip gives cos t = −, between −1 and 1.

Is the range of sec t the same as the range of cos t?

No.
cos t: [−1, 1].
sec t: (−∞, −1] ∪ [1, ∞).