Quotient Identities

Learn why tangent equals sine over cosine and cotangent equals cosine over sine, where they are defined, and how to use them to simplify expressions.

Advanced25 minLesson
Printable worksheet

Definition

The quotient identities write tangent and cotangent as a quotient (a division) of sine and cosine of the same angle:
tan⁡θ=sin⁡θcos⁡θ(cos⁡θ≠0)\tan\theta = \frac{\sin\theta}{\cos\theta} \qquad (\cos\theta \neq 0)
cot⁡θ=cos⁡θsin⁡θ(sin⁡θ≠0)\cot\theta = \frac{\cos\theta}{\sin\theta} \qquad (\sin\theta \neq 0)

Where they come from: the right triangle

For an acute angle θ\theta in a right triangle, sin⁡θ=oppositehypotenuse\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} and cos⁡θ=adjacenthypotenuse\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}. Divide them and the hypotenuse cancels:
sin⁡θcos⁡θ=opposite/hypotenuseadjacent/hypotenuse=oppositeadjacent=tan⁡θ\frac{\sin\theta}{\cos\theta} = \frac{\text{opposite}/\text{hypotenuse}}{\text{adjacent}/\text{hypotenuse}} = \frac{\text{opposite}}{\text{adjacent}} = \tan\theta

Where they come from: the unit circle

The angle θ\theta meets the unit circle at the point P=(cos⁡θ,sin⁡θ)P = (\cos\theta, \sin\theta). The slope of the radius OPOP is yx=sin⁡θcos⁡θ\frac{y}{x} = \frac{\sin\theta}{\cos\theta}, and that slope is tan⁡θ\tan\theta. This works for every angle, not only acute ones, so the signs come out right in all four quadrants.

Where they are defined

  • tan⁡θ\tan\theta is undefined when cos⁡θ=0\cos\theta = 0: at 90°90°, 270°270°, ... (that is θ=90°+k⋅180°\theta = 90° + k \cdot 180°).
  • cot⁡θ\cot\theta is undefined when sin⁡θ=0\sin\theta = 0: at 0°0°, 180°180°, 360°360°, ... (that is θ=k⋅180°\theta = k \cdot 180°).

Link to the reciprocal identities

From the reciprocal identities, cot⁡θ=1tan⁡θ\cot\theta = \frac{1}{\tan\theta}. Flipping sin⁡θcos⁡θ\frac{\sin\theta}{\cos\theta} gives cos⁡θsin⁡θ\frac{\cos\theta}{\sin\theta}, so both views agree, and tan⁡θ⋅cot⁡θ=1\tan\theta \cdot \cot\theta = 1 wherever both are defined.

Try it now

Which expression is equal to tan⁡θ\tan\theta?

Worked Examples

For an angle θ\theta we know sin⁡θ=35\sin\theta = \frac{3}{5} and cos⁡θ=45\cos\theta = \frac{4}{5}. Find tan⁡θ\tan\theta and cot⁡θ\cot\theta.
1

Write the quotient identity for tangent

tan⁡θ=sin⁡θcos⁡θ=3/54/5\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3/5}{4/5} → Divide the two fractions
2

Divide (the fifths cancel)

3/54/5=35⋅54=34\frac{3/5}{4/5} = \frac{3}{5} \cdot \frac{5}{4} = \frac{3}{4} → tan⁡θ=34\tan\theta = \frac{3}{4}
3

Use the quotient identity for cotangent

cot⁡θ=cos⁡θsin⁡θ=4/53/5=43\cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{4/5}{3/5} = \frac{4}{3} → cot⁡θ=43\cot\theta = \frac{4}{3}
4

Check with the reciprocal identity

tan⁡θ⋅cot⁡θ=34⋅43=1\tan\theta \cdot \cot\theta = \frac{3}{4} \cdot \frac{4}{3} = 1 → Correct

Common Mistakes

Writing tan⁡θ=cos⁡θsin⁡θ\tan\theta = \frac{\cos\theta}{\sin\theta}

Why it's wrong: The fraction is upside down: that is cotangent. On the unit circle, tangent is the slope yx\frac{y}{x}, and y=sin⁡θy = \sin\theta.

Correct: tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta} (sine on top, like yy over xx) and cot⁡θ=cos⁡θsin⁡θ\cot\theta = \frac{\cos\theta}{\sin\theta}

Saying tan⁡90°=0\tan 90° = 0 or cot⁡0°=0\cot 0° = 0

Why it's wrong: At 90°90° the cosine is 00, and you cannot divide by 00. At 0°0° the sine is 00, so cotangent has a zero denominator.

Correct: tan⁡90°=10\tan 90° = \frac{1}{0} is undefined, and cot⁡0°=10\cot 0° = \frac{1}{0} is undefined. The graph of tan⁡\tan has vertical asymptotes there.

Mixing two angles: sin⁡40°cos⁡20°=tan⁡20°\frac{\sin 40°}{\cos 20°} = \tan 20° or tan⁡40°\tan 40°

Why it's wrong: The identity only works when sine and cosine have the same angle.

Correct: sin⁡40°cos⁡40°=tan⁡40°\frac{\sin 40°}{\cos 40°} = \tan 40°, but sin⁡40°cos⁡20°≈0.684\frac{\sin 40°}{\cos 20°} \approx 0.684 is neither tan⁡20°≈0.364\tan 20° \approx 0.364 nor tan⁡40°≈0.839\tan 40° \approx 0.839.

Losing the sign in quadrants II, III and IV

Why it's wrong: Students often use the positive square root for cos⁡θ\cos\theta or sin⁡θ\sin\theta and get tan⁡θ>0\tan\theta > 0 everywhere.

Correct: Take the signs from the quadrant first: in quadrant II sin⁡>0\sin > 0, cos⁡<0\cos < 0, so tan⁡θ<0\tan\theta < 0; in quadrant III both are negative, so tan⁡θ>0\tan\theta > 0.

Interactive Visual

Prove

tan⁡θcos⁡θ=sin⁡θ\tan\theta \cos\theta = \sin\theta

  1. Start with the left side

    tan⁡θcos⁡θ\tan\theta \cos\theta

Check with numbers

Left side

tan⁡θcos⁡θ\tan\theta \cos\theta

= 0.5

Right side

sin⁡θ\sin\theta

= 0.5

Both sides have the same value.

Function graph

-300-200-1000100200300-4-2024xyx = -270x = -90x = 90x = 270x = -270x = -90x = 90x = 270

Tap or hover over the graph to read off values.

  • sin(a) = 0.5
  • cos(a) = 0.87
  • sin(a)/cos(a) = 0.58
  • tan(a) = 0.58
  • cos(a)/sin(a) = 1.73
  • cot(a) = 1.73

Tap a name to show or hide its graph.

Asymptote

Unit Circle

Degrees

0°

Radians

0

sin(θ)

0

cos(θ)

1

tan(θ)

0

Coordinates (cos, sin)

(1, 0)

Click and drag to rotate the angle around the unit circle.

Interactive Sandbox

Expression Calculator

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Practice Problems

18 problems
Problem 1 of 18
Easy
Which expression is equal to tan⁡θ\tan\theta?

Why It Matters

The quotient identities are the bridge that turns every trigonometric expression into sines and cosines:
  • Simplifying: rewriting tan⁡\tan and cot⁡\cot as fractions of sin⁡\sin and cos⁡\cos is the first move in almost every simplification and proof that follows in this topic
  • Exact values: if you know sin⁡θ\sin\theta and cos⁡θ\cos\theta (from the unit circle or a triangle), you get tan⁡θ\tan\theta and cot⁡θ\cot\theta with one division
  • Slopes and angles: the slope of a line, a road or a ramp is the tangent of its angle of inclination
  • Physics: an object on a ramp starts to slide when the friction coefficient is smaller than tan⁡θ\tan\theta, and shadow lengths use cot⁡θ\cot\theta
  • Calculus: the derivative of tan⁡x\tan x is found by writing it as sin⁡xcos⁡x\frac{\sin x}{\cos x} and using the quotient rule

Real World Applications

Ski Slopes and Road Gradients

The slope of a hill (rise over horizontal run) is the tangent of its angle of inclination. A sensor in a phone measures sine and cosine of the tilt, and the app divides them.

Example:

A slope with sin⁡θ=0.6\sin\theta = 0.6 and cos⁡θ=0.8\cos\theta = 0.8 has tan⁡θ=0.75\tan\theta = 0.75: it drops 7575 m for every 100100 m of horizontal distance.

1Try It Yourself

A tilt sensor on a ski slope reads sin⁡θ=0.6\sin\theta = 0.6 and cos⁡θ=0.8\cos\theta = 0.8. The run covers 200200 m of horizontal distance.

How many meters does the run drop in height?

Step 1: Write the mathematical expression

Height drop = tan⁡θ×\tan\theta \times horizontal distance:

Friction on a Ramp

A box on a ramp stays at rest as long as the friction coefficient $\mu$ is at least $\tan\theta$. Engineers use this to decide how steep conveyor belts and loading ramps may be.

Example:

On a ramp with tan⁡θ=0.5\tan\theta = 0.5, a box with μ=0.6\mu = 0.6 stays put, but a box with μ=0.4\mu = 0.4 slides.

2Try It Yourself

A loading ramp has sin⁡θ=941\sin\theta = \frac{9}{41} and cos⁡θ=4041\cos\theta = \frac{40}{41}. A plastic crate on it has friction coefficient μ=0.2\mu = 0.2.

What is the smallest friction coefficient that keeps a crate at rest on this ramp?

Step 1: Write the mathematical expression

Smallest μ\mu = tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}:

Shadow Lengths

When the sun stands at an elevation angle $\theta$, a pole of height $h$ casts a shadow of length $h \cot\theta = h \cdot \frac{\cos\theta}{\sin\theta}$. Architects use this to plan shade and sunlight.

Example:

With the sun at 45°45°, cot⁡45°=1\cot 45° = 1, so every object casts a shadow as long as it is tall.

3Try It Yourself

At noon the sun's elevation angle has sin⁡θ=0.96\sin\theta = 0.96 and cos⁡θ=0.28\cos\theta = 0.28. A flagpole is 1212 m tall.

How long is the flagpole's shadow, in meters?

Step 1: Write the mathematical expression

Shadow = 12⋅cos⁡θsin⁡θ12 \cdot \frac{\cos\theta}{\sin\theta}:

Key Takeaways

  • 1tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta} and cot⁡θ=cos⁡θsin⁡θ\cot\theta = \frac{\cos\theta}{\sin\theta}, always for the same angle θ\theta
  • 2In a right triangle the hypotenuse cancels: sin⁡θcos⁡θ=oppositeadjacent\frac{\sin\theta}{\cos\theta} = \frac{\text{opposite}}{\text{adjacent}}
  • 3On the unit circle, tan⁡θ\tan\theta is the slope yx\frac{y}{x} of the radius to (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta), so it works in every quadrant
  • 4tan⁡θ\tan\theta is undefined where cos⁡θ=0\cos\theta = 0 (90°90°, 270°270°, ...); cot⁡θ\cot\theta is undefined where sin⁡θ=0\sin\theta = 0 (0°0°, 180°180°, ...)
  • 5tan⁡θ⋅cot⁡θ=1\tan\theta \cdot \cot\theta = 1 wherever both are defined
  • 6To simplify, rewrite everything in sin⁡\sin and cos⁡\cos first, then cancel common factors

Frequently Asked Questions

A quotient is the result of a division. These identities express tan⁡\tan and cot⁡\cot as one function divided by another: sin⁡θ÷cos⁡θ\sin\theta \div \cos\theta and cos⁡θ÷sin⁡θ\cos\theta \div \sin\theta.
A quotient is the result of a division. These identities express tan⁡\tan and cot⁡\cot as one function divided by another: sin⁡θ÷cos⁡θ\sin\theta \div \cos\theta and cos⁡θ÷sin⁡θ\cos\theta \div \sin\theta.
Both views lead to the same place. The reciprocal identity says cot⁡θ=1tan⁡θ\cot\theta = \frac{1}{\tan\theta}; replacing tan⁡θ\tan\theta with sin⁡θcos⁡θ\frac{\sin\theta}{\cos\theta} and flipping gives the quotient form cos⁡θsin⁡θ\frac{\cos\theta}{\sin\theta}. The quotient form also works at 90°90°, where tan⁡\tan is undefined but cot⁡90°=01=0\cot 90° = \frac{0}{1} = 0.
Because they are the same function. Both have vertical asymptotes where cos⁡x=0\cos x = 0 (x=±90°,±270°,...x = \pm 90°, \pm 270°, ...) and zeros where sin⁡x=0\sin x = 0. Toggle the two curves in the graph above: they lie exactly on top of each other.
Yes. Type sin⁡(40)/cos⁡(40)\sin(40) / \cos(40) and tan⁡(40)\tan(40) in degree mode: both give about 0.83910.8391. A numeric check at a few angles is a good test, but only an algebraic argument proves an identity for all angles.

Glossary

Quotient
The result of a division, for example ab\frac{a}{b} is the quotient of aa and bb
Quotient identities
tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta} and cot⁡θ=cos⁡θsin⁡θ\cot\theta = \frac{\cos\theta}{\sin\theta}
Identity
An equation that is true for every value of the variable for which both sides are defined
Cotangent (cot⁡\cot)
The ratio cos⁡θsin⁡θ\frac{\cos\theta}{\sin\theta}, the reciprocal of tangent
Domain
The set of angles where an expression is defined; for tan⁡θ\tan\theta all angles with cos⁡θ≠0\cos\theta \neq 0

Formula Card

Tangent

tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}

Defined when $\cos\theta \neq 0$

Cotangent

cot⁡θ=cos⁡θsin⁡θ\cot\theta = \frac{\cos\theta}{\sin\theta}

Defined when $\sin\theta \neq 0$

Product

tan⁡θ⋅cot⁡θ=1\tan\theta \cdot \cot\theta = 1

Where both are defined

Right triangle

tan⁡θ=oppositeadjacent\tan\theta = \frac{\text{opposite}}{\text{adjacent}}

The hypotenuse cancels in $\frac{\sin\theta}{\cos\theta}$

Today's challengeOne problem, three tries

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