Secant, Cosecant, and Cotangent

Understanding the reciprocal trigonometric functions and their graphs

SPEC 5.4 • TRIGONOMETRY

Learning Objectives

By the end of this lesson

Understand the definitions of

sec θ

,

csc θ

, and

cot θ

Know the relationships between reciprocal and primary trig functions

Evaluate exact values of reciprocal functions at key angles

Sketch graphs of

y = sec x

,

y = csc x

, and

y = cot x

Identify domains, ranges, and asymptotes of reciprocal functions

Prior Knowledge

Graphs of

sin x

,

cos x

, and

tan x

; exact trigonometric values; understanding of asymptotes.

The Three Reciprocal Functions

Just as division is the inverse operation of multiplication, each primary trigonometric function has a reciprocal function.

The Reciprocal Functions

sec θ

Secant

reciprocal of cos

csc θ

Cosecant

reciprocal of sin

cot θ

Cotangent

reciprocal of tan

Notation Note

Cosecant may be written as

csc θ

or

cosec θ

. Both are acceptable.

Secant: Definition and Graph

Definition

$$\sec\theta = \frac{1}{\cos\theta}$$

Period

2 π

Domain

x \neq = \frac{π}{2} + nπ

Range

y \leq - 1

or

y \geq 1

Memory Aid

SECant → Cosine (3rd letter of "sec" is C). Asymptotes where

cos x = 0

.

Cosecant: Definition and Graph

Definition

$$\csc\theta = \frac{1}{\sin\theta}$$

Period

2 π

Domain

x \neq = nπ

Range

y \leq - 1

or

y \geq 1

Memory Aid

COsecant → Sine (the "co-" pairs with sine). Asymptotes where

sin x = 0

.

Cotangent: Definition and Graph

Definition

$$\cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}$$

Period

π

Domain

x \neq = nπ

Range

y \in R

(all reals)

Key Feature

Unlike sec and csc, cotangent is a decreasing function within each period. Crosses the x-axis at

x = \frac{π}{2} + nπ

.

Critical Distinction

Common Mistake

Reciprocal ≠ Inverse

RECIPROCAL

sec θ = \frac{1}{cos θ}

"One divided by cos θ"

Output is a number

INVERSE

cos^{- 1} x = arccos (x)

"The angle whose cosine is x"

Output is an angle

Remember

sec θ \neq = cos^{- 1} θ

— these are completely different!

csc θ \neq = sin^{- 1} θ

cot θ \neq = tan^{- 1} θ

Exact Values Table

Using our knowledge of exact values for sin, cos, and tan:

θ

0

\frac{π}{6}

\frac{π}{4}

\frac{π}{3}

\frac{π}{2}

sec θ

1

\frac{2 3}{3}

2

2Undefined

csc θ

Undefined2

2

\frac{2 3}{3}

1

cot θ

Undefined

3

1

\frac{3}{3}

0

Method

These values are found by taking reciprocals:

sec \frac{π}{3} = \frac{1}{c o s \frac{π}{3}} = \frac{1}{1/2} = 2

Worked Example01

Find the exact values of: (a) $sec 60°$ and (b) $csc 45°$

a)

Recall

cos 60° = \frac{1}{2}

•

sec 60° = \frac{1}{cos 60°} = \frac{1}{1/2} = 2

b)

Recall

sin 45° = \frac{2}{2}

•

csc 45° = \frac{1}{sin 45°} = \frac{1}{\frac{2}{2}} = \frac{2}{2} = 2

Answer(a)

sec 60° = 2

(b)

csc 45° = 2

Worked Example02

Find the exact value of $cot (\frac{5 π}{6})$

1

\frac{5 π}{6}

is in the second quadrant (cos negative, sin positive)

2

Reference angle:

π - \frac{5 π}{6} = \frac{π}{6}

3

cos \frac{5 π}{6} = - \frac{3}{2}

and

sin \frac{5 π}{6} = \frac{1}{2}

4

cot (\frac{5 π}{6}) = \frac{cos \frac{5 π}{6}}{sin \frac{5 π}{6}} = \frac{- \frac{3}{2}}{\frac{1}{2}} = - 3

Answer

cot (\frac{5 π}{6}) = - 3

Worked Example03

Given that $sin θ = \frac{3}{5}$ and $θ$ is acute, find the exact values of $sec θ$ and $cot θ$ .

1

Using

sin^{2} θ + cos^{2} θ = 1

:

cos^{2} θ = 1 - \frac{9}{25} = \frac{16}{25}

2

Since

θ

is acute:

cos θ = \frac{4}{5}

3

sec θ = \frac{1}{cos θ} = \frac{1}{4/5} = \frac{5}{4}

4

cot θ = \frac{cos θ}{sin θ} = \frac{4/5}{3/5} = \frac{4}{3}

Answer

sec θ = \frac{5}{4}

,

cot θ = \frac{4}{3}

Worked Example04

Simplify $sec θ \cdot sin θ$

1

Replace

sec θ

with its definition

2

sec θ \cdot sin θ = \frac{1}{cos θ} \cdot sin θ

3

= \frac{sin θ}{cos θ}

4

Recognise this as

tan θ

Answer

sec θ \cdot sin θ = tan θ

P

Practice 1

Find the exact value of each:

(a)

sec (\frac{π}{4})

(b)

csc (\frac{π}{3})

(c)

cot (\frac{3 π}{4})

a)

sec \frac{π}{4} = \frac{1}{c o s \frac{π}{4}} = \frac{1}{\frac{2}{2}} = 2

b)

csc \frac{π}{3} = \frac{1}{s i n \frac{π}{3}} = \frac{1}{\frac{3}{2}} = \frac{2}{3} = \frac{2 3}{3}

c)

cot \frac{3 π}{4} = \frac{c o s \frac{3 π}{4}}{s i n \frac{3 π}{4}} = \frac{- \frac{2}{2}}{\frac{2}{2}} = - 1

P

Practice 2

Given that $cos θ = - \frac{5}{13}$ and $\frac{π}{2} < θ < π$ , find:

(a)

sec θ

(b)

csc θ

(c)

cot θ

•

First find

sin θ

:

sin^{2} θ = 1 - \frac{25}{169} = \frac{144}{169}

•

In Q2,

sin θ > 0

, so

sin θ = \frac{12}{13}

a)

sec θ = \frac{1}{c o s θ} = \frac{1}{- \frac{5}{13}} = - \frac{13}{5}

b)

csc θ = \frac{1}{s i n θ} = \frac{1}{\frac{12}{13}} = \frac{13}{12}

c)

cot θ = \frac{c o s θ}{s i n θ} = \frac{- \frac{5}{13}}{\frac{12}{13}} = - \frac{5}{12}

Summary

Reciprocal Function Definitions

sec θ = \frac{1}{cos θ}

csc θ = \frac{1}{sin θ}

cot θ = \frac{1}{tan θ} = \frac{cos θ}{sin θ}

Key Graph Properties

FunctionPeriodRangeAsymptotes at

y = sec x

2 π

y \leq - 1

or

y \geq 1

x = \frac{π}{2} + nπ

y = csc x

2 π

y \leq - 1

or

y \geq 1

x = nπ

y = cot x

π

y \in R

x = nπ

Remember

Reciprocal ≠ Inverse!

sec θ \neq = cos^{- 1} θ