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Sec squared and cosec squared identities

Derive 1 + tan²x = sec²x and 1 + cot²x = cosec²x. Find exact signed ratios, prove higher-power identities and eliminate a trigonometric parameter.

Before you startSine, cosine, reciprocal trig functions and Pythagoras

01 / Two identities from the same circle

Divide Pythagoras by a nonzero squared coordinate.

The familiar identity sin²θ + cos²θ = 1 produces two more useful identities. Divide every term by cos²θ to connect tangent and secant, or by sin²θ to connect cotangent and cosecant.

1 + tan²θ = sec²θ, if cos θ ≠ 0
1 + cot²θ = cosec²θ, if sin θ ≠ 0

The model shows a normalised right triangle and the matching squared lengths. In other quadrants the lengths use magnitudes; the signed trig values appear separately below the diagram.

One squared leg plus the otherExplore

tan θ = 1.732; sec θ = 2.000.

1 + 3.000 = 4.000.

Requires cos θ ≠ 0. This input is allowed.

Green area: 1. Blue area: the other leg squared. Gold area: the hypotenuse squared. Values are shown to 3 decimal places. Axis angles may give a degenerate triangle while the algebra remains valid.

IdentityAngle

Watch a unit-circle triangle rescale

Pause, replay or seek freely. The notes explain the same idea and stay in view.

02 / Derive both formulas term by term

The same denominator must divide every term.

Divide by cos²θWorked example

sin²θ/cos²θ + cos²θ/cos²θ = 1/cos²θ

Require cos θ ≠ 0.

tan²θ + 1 = sec²θ

Use the ratio and reciprocal definitions.

Divide by sin²θWorked example

sin²θ/sin²θ + cos²θ/sin²θ = 1/sin²θ

Require sin θ ≠ 0.

1 + cot²θ = cosec²θ

The cot term is cosine squared over sine squared.

At θ = 0, the tangent/secant identity is valid: 1 +0 =1. The cotangent/cosecant identity is undefined there. At θ = π/2 their roles reverse.

03 / Choose the useful rearrangement

Check which squared function the question needs.

tan²θ = sec²θ − 1
cot²θ = cosec²θ − 1
sec²θ − tan²θ = 1
cosec²θ − cot²θ = 1

If sec²θ = 13/4, find tan²θWorked example

tan²θ = 13/4 −1 = 9/4

Subtract the whole number 1, written as 4/4.

tan θ = ±3/2

The squared value alone does not choose the sign.

The identity is a sum of squares. It does not imply sec θ = 1 + tan θ or cosec θ = 1 + cot θ.

04 / Choose square-root signs from the quadrant

A negative secant value is perfectly possible.

Given tan θ = −7/24 and π/2 < θ < π, find sec θ, cosec θ and cot θWorked example

sec²θ = 1 +49/576 = 625/576

Use the tangent/secant identity.

sec θ = −25/24

In Quadrant II cosine, and therefore secant, is negative.

cot θ = −24/7

Tangent is nonzero, so take its reciprocal.

cosec²θ = 1 +576/49 = 625/49

Use the cotangent/cosecant identity.

cosec θ = 25/7

Sine is positive in Quadrant II.

Keep the quadrant statement beside your work. A root sign chosen from habit can reverse the answer even when the squared calculation is correct.

05 / Keep sign information in parameter formulas

Restrictions on an angle also restrict the parameter.

Let tan θ = p with π < θ < 3π/2Worked example

Here p > 0

Tangent is positive in Quadrant III.

sec²θ = 1 + p², so sec θ = −√(1 + p²)

Cosine is negative.

cot θ = 1/p

The strict quadrant ensures p ≠ 0.

cosec θ = −√(1 + p²)/p

Its square is 1 +1/p² and its sign is negative.

Let cos θ = k with π/2 < θ < πWorked example

−1 < k < 0

The angle lies strictly inside Quadrant II.

sin θ = √(1 −k²)

Use the positive root.

sec θ = 1/k; cosec θ = 1/√(1 −k²)

Neither denominator vanishes under the strict bounds.

cot θ = k/√(1 −k²)

This is negative as required.

06 / Factor higher powers before substituting

A difference of fourth powers hides a difference of squares.

Prove sec⁴θ − tan⁴θ = 1 + 2tan²θWorked example

= (sec²θ −tan²θ)(sec²θ +tan²θ)

Factor the difference of two squares.

= 1 × (sec²θ +tan²θ)

Use sec²θ −tan²θ =1.

= 1 +2tan²θ

Replace the remaining sec²θ by 1 +tan²θ. Require cos θ ≠0.

Find an alternative form for cosec⁴θ − cot⁴θWorked example

= (cosec²θ −cot²θ)(cosec²θ +cot²θ)

The first factor is 1.

= 1 +2cot²θ = 2cosec²θ −1

Both forms are useful, with sin θ ≠0.

Changing everything into sine and cosine remains valid, but spotting a factor of 1 can make the proof much shorter.

07 / Factor a rational expression and retain its exclusions

An extra denominator can remove additional inputs.

Simplify tan²θ/(sec θ +1)Worked example

Original conditions: cos θ ≠0 and sec θ ≠−1

The sec and tan functions must exist; the outer denominator must also be nonzero.

Replace tan²θ by sec²θ −1

Factor the numerator as (sec θ −1)(sec θ +1).

Cancel the nonzero factor sec θ +1

The result is sec θ −1.

Keep the original conditions

Odd multiples of π are still excluded, even though the simplified expression has value −2 there.

Prove (sec θ +tan θ)² = (1 +sin θ)/(1 −sin θ) on their common domainWorked example

LHS = (1 +sin θ)²/cos²θ

Use the definitions, requiring cos θ ≠0.

= (1 +sin θ)²/((1 −sin θ)(1 +sin θ))

Factor 1 −sin²θ.

= (1 +sin θ)/(1 −sin θ)

Since cosine is nonzero, sine is neither1 nor−1, so the cancellation is valid.

The right expression alone exists at sin θ = −1, while the left does not. The common domain still excludes every zero of cosine.

08 / Eliminate a trigonometric parameter

A squared relation may describe more points than the original angle range.

x = 3tan θ and y = 5sec θWorked example

tan θ = x/3 and sec θ = y/5

Express the two linked functions in terms of x and y.

1 + x²/9 = y²/25

Substitute in the tangent/secant identity.

y²/25 −x²/9 =1

This is the Cartesian relation.

If 0 < θ < π/2, keep x >0 and y >5

The squared relation alone also includes negative y and other x-values.

Without an additional angle restriction, the sec/tan parameterisation covers both branches with |y| ≥5. On the stated acute interval it covers only the part in the first quadrant with strict bounds.

09 / Check whether proposed values can coexist

The identities test consistency, but signs and domains still matter.

Could tan θ = 3/4 and sec θ = −5/4 describe the same real angle?Worked example

1 +(3/4)² =25/16 =(−5/4)²

The squared identity is satisfied.

Tangent is positive and secant negative

Cosine is negative, so sine must also be negative.

Given sec θ +tan θ =4, find sec θ and tan θ.

Hint

The product of sec+tan and sec−tan is 1.

Worked solution

Sec θ −tan θ =1/4. Adding and subtracting gives sec θ =17/8 and tan θ =15/8. Check(17/8)² −(15/8)² =1; both signs are consistent with Quadrant I.

11 / Recap

A squared identity supplies magnitudes; other information supplies signs.

Divide sin²θ +cos²θ =1 by a nonzero squared denominator.
1 +tan²θ =sec²θ and 1 +cot²θ =cosec²θ.
Choose root signs from the quadrant.
Factor differences of higher powers before substituting.
Retain all original exclusions when cancelling fractions.
Parameter elimination can add points unless angle restrictions are carried forward.

Section 1 of 11 · Two identities from the same circle