Trigonometric Identities

The ratio identity and Pythagorean identities

SPEC 5.6 • TRIGONOMETRIC IDENTITIES

What is an Identity?

Definition

An identity is an equation that is true for all values of the variable (within its domain).

We use the symbol ≡ (triple equals) to show an identity, rather than = (single equals).

Identity

2 (x + 3) \equiv 2 x + 6

True for ALL values of x

Equation

2 x + 6 = 10

True only when x = 2

Note

In trigonometry, θ (theta) represents any angle. We could equally use x, α (alpha), β (beta), or any other letter. The identities work for all angle values.

The Ratio Identity

From the definitions of sine, cosine, and tangent:

tan θ = \frac{opposite}{adjacent} = \frac{opposite / hypotenuse}{adjacent / hypotenuse} = \frac{sin θ}{cos θ}

Ratio Identity

tan θ \equiv \frac{sin θ}{cos θ}

Restriction

This identity is valid only where

cos θ \neq = 0

, i.e.,

θ \neq = 90°, 270°, ...

or

θ \neq = \frac{π}{2} + nπ

Remember

This identity is essential for simplifying expressions and converting between trig functions.

The Fundamental Pythagorean Identity

Consider a point P on the unit circle (radius = 1) at angle θ:

The coordinates of P are

(cos θ, sin θ)

. Since P lies on the unit circle:

x^{2} + y^{2} = 1 \Rightarrow cos^{2} θ + sin^{2} θ = 1

First Pythagorean Identity

sin^{2} θ + cos^{2} θ \equiv 1

Important Notation

sin²θ Notation

sin^{2} θ means (sin θ)^{2}

It does NOT mean

sin (θ^{2})

Correct

sin^{2} 30° = (sin 30°)^{2} = (0.5)^{2} = 0.25

Incorrect

sin^{2} 30° \neq = sin (900°)

Same Pattern

This notation applies to all trig functions:

cos^{2} θ = (cos θ)^{2}

,

tan^{2} θ = (tan θ)^{2}

, etc.

The Second Pythagorean Identity

Starting from:

sin^{2} θ + cos^{2} θ = 1

Divide every term by $cos^{2} θ$ :

\frac{sin ^{2} θ}{cos ^{2} θ} + \frac{cos ^{2} θ}{cos ^{2} θ} = \frac{1}{cos ^{2} θ}

tan^{2} θ + 1 = sec^{2} θ

Second Pythagorean Identity

1 + tan^{2} θ \equiv sec^{2} θ

Restriction

Valid only where

cos θ \neq = 0

The Third Pythagorean Identity

Starting from:

sin^{2} θ + cos^{2} θ = 1

Divide every term by $sin^{2} θ$ :

\frac{sin ^{2} θ}{sin ^{2} θ} + \frac{cos ^{2} θ}{sin ^{2} θ} = \frac{1}{sin ^{2} θ}

1 + cot^{2} θ = csc^{2} θ

Third Pythagorean Identity

1 + cot^{2} θ \equiv csc^{2} θ

Restriction

Valid only where

sin θ \neq = 0

Summary: Key Trigonometric Identities

RATIO IDENTITY

tan θ \equiv \frac{sin θ}{cos θ}

PYTHAGOREAN 1

sin^{2} θ + cos^{2} θ \equiv 1

PYTHAGOREAN 2

1 + tan^{2} θ \equiv sec^{2} θ

PYTHAGOREAN 3

1 + cot^{2} θ \equiv csc^{2} θ

Memory Tip

Notice the pattern: each identity pairs a squared trig function with its corresponding reciprocal squared. The "co" functions (cot, csc) go together, and the non-"co" functions (tan, sec) go together.

Worked Example01

Simplify:

\frac{1 - sin ^{2} x}{cos x}

1

Since

sin^{2} x + cos^{2} x = 1

, we have

1 - sin^{2} x = cos^{2} x

2

Substitute:

\frac{cos ^{2} x}{cos x}

3

Cancel:

cos x

\frac{1 - sin ^{2} x}{cos x} = cos x

Worked Example02

Simplify:

sin α \cdot cot α

1

Write cot α in terms of sin and cos:

cot α = \frac{cos α}{sin α}

2

Substitute:

sin α \cdot \frac{cos α}{sin α}

3

Cancel sin α:

cos α

sin α \cdot cot α = cos α

P1

Practice Problem

Simplify: $cos β \cdot tan β$

1

tan β = \frac{sin β}{cos β}

2

cos β \cdot \frac{sin β}{cos β} = sin β

cos β \cdot tan β = sin β

Worked Example03

Simplify:

\frac{sin ^{2} θ}{cos ^{2} θ} + 1

1

Recognize that

\frac{sin ^{2} θ}{cos ^{2} θ} = tan^{2} θ

2

So we have:

tan^{2} θ + 1

3

Using the identity

1 + tan^{2} θ = sec^{2} θ

:

\frac{sin ^{2} θ}{cos ^{2} θ} + 1 = sec^{2} θ

Worked Example04

Simplify:

sec^{2} A - tan^{2} A

1

From the identity

1 + tan^{2} θ = sec^{2} θ

2

Rearranging:

sec^{2} θ - tan^{2} θ = 1

3

This works for any angle, including A:

sec^{2} A - tan^{2} A = 1

P2

Practice Problem

Simplify: $csc^{2} ϕ - cot^{2} ϕ$

1

From

1 + cot^{2} θ = csc^{2} θ

2

Rearranging:

csc^{2} θ - cot^{2} θ = 1

csc^{2} ϕ - cot^{2} ϕ = 1

Worked Example05

Simplify:

tan θ \cdot csc θ

1

Convert to sin and cos:

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

2

Simplify:

\frac{sin θ}{cos θ \cdot sin θ}

3

Cancel sin θ:

\frac{1}{cos θ}

tan θ \cdot csc θ = sec θ

Worked Example06

Simplify:

sec^{4} x - tan^{4} x

1

Factor as difference of squares:

(sec^{2} x + tan^{2} x) (sec^{2} x - tan^{2} x)

2

We know

sec^{2} x - tan^{2} x = 1

3

So:

(sec^{2} x + tan^{2} x) (1) = sec^{2} x + tan^{2} x

sec^{4} x - tan^{4} x = sec^{2} x + tan^{2} x

P3

Practice Problem

Simplify:

cot θ \cdot sec θ \cdot sin θ

1

Convert:

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

2

Simplify:

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

cot θ \cdot sec θ \cdot sin θ = 1

Worked Example07

Simplify:

\frac{sin ^{2} α + cos ^{2} α}{sec ^{2} α - tan ^{2} α}

1

The numerator:

sin^{2} α + cos^{2} α = 1

2

The denominator:

sec^{2} α - tan^{2} α = 1

3

Therefore:

\frac{1}{1} = 1

\frac{sin ^{2} α + cos ^{2} α}{sec ^{2} α - tan ^{2} α} = 1

P4

Practice Problem

Simplify:

\frac{1 - cos ^{2} θ}{sin θ}

1

Since

sin^{2} θ + cos^{2} θ = 1

, we have

1 - cos^{2} θ = sin^{2} θ

2

So:

\frac{sin ^{2} θ}{sin θ} = sin θ

\frac{1 - cos ^{2} θ}{sin θ} = sin θ

Lesson Summary

Key Identities to Remember

tan θ \equiv \frac{sin θ}{cos θ}

sin^{2} θ + cos^{2} θ \equiv 1

1 + tan^{2} θ \equiv sec^{2} θ

1 + cot^{2} θ \equiv csc^{2} θ

Simplifying Strategy

Convert everything to sin and cos if unsure

Look for

sin^{2} + cos^{2}

to replace with 1

Look for

sec^{2} - tan^{2}

or

csc^{2} - cot^{2}

to replace with 1

Use algebraic techniques like factoring (difference of squares)

Remember

The angle can be θ, x, α, β, A, or any other variable — the identities work the same way for all angles!