01 · Cancel a factor
Simplify cos θ sec θ.
Hint
Replace sec by 1/cos.
Worked solution
The result is 1, with cos θ ≠ 0. The original product is undefined at odd multiples of π/2.
Understand · explore · practise
Reciprocal trigonometric identities
Prove identities involving sec, cosec and cot by rewriting in sine and cosine. Keep domain restrictions, handle square roots and apply the algebra to geometry.
Before you startReciprocal trig definitions; algebraic fractions; sin²θ + cos²θ = 1
01 / Start with the definitions and the domain
A proof explains why two expressions agree for every permitted input.
To prove a reciprocal trigonometric identity, rewrite sec, cosec and cot in terms of sine and cosine. Then use ordinary fraction algebra and sin²θ + cos²θ = 1. Record the original exclusions before cancelling anything.
sec θ = 1/cos θ
cosec θ = 1/sin θ
cot θ = cos θ/sin θ
The model compares values at selected angles. It also shows when an original expression is undefined even though its simplified form has a value. Matching a few numerical examples is a check, not a proof for all angles.
Equal values, with an input checkExplore
sin θ cot θ = cos θ
LHS = 0.500; RHS = 0.500.
Original domain: sin θ ≠ 0. This input is allowed.
Blue: original left side. Green: simplified right side. A missing point is undefined. A numerical match suggests a pattern; the algebra proves the identity.
Watch cancellation leave the excluded inputs behind
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Cancel factors only after rewriting
Keep a note of what was required to be nonzero.
Simplify sin θ cot θWorked example
sin θ × (cos θ/sin θ)
The original cot requires sin θ ≠ 0.
= cos θ
Cancel the common nonzero factor sin θ.
The identity holds for θ ≠ nπ
The simpler right side also exists at nπ, but the original left side does not.
Simplify sec³θ cos⁴θWorked example
(1/cos³θ) × cos⁴θ = cos θ
The original sec requires cos θ ≠ 0.
Keep θ ≠ π/2 + nπ
Cancelling three powers does not make the original expression defined there.
Cancel multiplied factors, not pieces of a sum. For example, you cannot cancel the “cos” terms across sec θ + cos θ.
03 / Use a common denominator
A sum of squares often produces the number 1.
Prove tan θ + cot θ = sec θ cosec θWorked example
tan θ + cot θ = sin θ/cos θ + cos θ/sin θ
Both sine and cosine must be nonzero.
= (sin²θ + cos²θ)/(sin θ cos θ)
Use the product as a common denominator.
= 1/(sin θ cos θ) = sec θ cosec θ
The Pythagorean identity simplifies the numerator.
Simplify 2sin θ cot θ + 3cos θWorked example
= 2cos θ + 3cos θ
Use the product identity, retaining sin θ ≠ 0.
= 5cos θ
Only combine like terms after simplifying the trig product.
04 / Turn a difference into a product
The identity can be useful in either direction.
Prove sec θ − cos θ = sin θ tan θWorked example
sec θ − cos θ = 1/cos θ − cos θ
Require cos θ ≠ 0.
= (1 − cos²θ)/cos θ
Write the terms over a common denominator.
= sin²θ/cos θ = sin θ tan θ
Replace 1 − cos²θ by sin²θ.
Simplify (sec θ − cos θ)/tan θWorked example
The original needs cos θ ≠ 0 and tan θ ≠ 0
Equivalently, both cosine and sine must be nonzero.
Replace the numerator by sin θ tan θ
Use the proved identity on its domain.
Cancel tan θ to obtain sin θ
Keep both original exclusions, even though sine itself is defined everywhere.
Multiplying a true identity by another expression or dividing it by a factor can change the allowed domain. Check that new operation separately.
05 / Prove a fractional identity from one side
Start with the side that contains more structure.
Prove (1 + cos θ)/sin θ = cosec θ + cot θWorked example
(1 + cos θ)/sin θ = 1/sin θ + cos θ/sin θ
Split the numerator term by term. Require sin θ ≠ 0.
= cosec θ + cot θ
Use the definitions directly.
Show (cosec θ + cot θ)(cosec θ − cot θ) = 1Worked example
= ((1 + cos θ)/sin θ)((1 − cos θ)/sin θ)
Require sin θ ≠ 0.
= (1 − cos²θ)/sin²θ
Use the difference of two squares.
= sin²θ/sin²θ = 1
The denominator is nonzero on the original domain.
This proves that the two factors are reciprocals wherever they exist. In a proof, transform one side into the other with valid steps; do not assume the desired equality and simply rearrange it until you see a true statement.
06 / Keep powers and square roots distinct
The square root gives a nonnegative result.
Simplify √(sec²θ)Worked example
√(z²) = |z| for real z
The principal square root is never negative.
Therefore √(sec²θ) = |sec θ|
Keep cos θ ≠ 0.
At θ = 2π/3, sec θ = −2 but √(sec²θ) = 2
Dropping the absolute value would change the result.
Simplify √(cosec²θ cot²θ)Worked example
The expression inside the root is (cosec θ cot θ)²
Both functions require sin θ ≠ 0.
Answer: |cosec θ cot θ|
It is not always cosec θ cot θ.
Since cosec θ cot θ = cos θ/sin²θ, its sign is the sign of cosine
If the angle is known to be acute, the absolute value may be removed. Otherwise it stays.
07 / Use the identity in a geometric calculation
Subtract two lengths along the same ray.
Triangle OAT is right angled at A, OA = d, and ∠AOT = θ is acute. Drop the perpendicular from A to OT, meeting it at B. Both OB and OT are measured along the same line.
Find BT in two equivalent formsWorked example
OT = d/cos θ = d sec θ
In OAT, OA is adjacent to θ and OT is the hypotenuse.
OB = d cos θ
In OAB, OA is the hypotenuse.
BT = OT − OB = d(sec θ − cos θ)
B lies between O and T for an acute angle.
= d sin θ tan θ
Use the difference identity.
Take d = 8 and θ = π/3Worked example
OT = 16 and OB = 4
Use sec(π/3) = 2 and cos(π/3) = 1/2.
BT = 12
The product form gives 8(√3/2)(√3) = 12 as well.
The equality of the two answers is a check; the algebraic proof above is what makes the formula valid for every acute θ.
08 / Audit a proposed proof
An algebraic step can be correct while its domain statement is wrong.
“sin θ cot θ = cos θ, so it is valid at θ = 0.”Worked example
The cancellation assumed sin θ ≠ 0
At θ = 0 the original cot contains division by 0.
The right side has value 1, but the left side is undefined
The simplified expression extends the formula to a larger domain.
Correct statement: the identity holds for θ ≠ nπ
The holes belong to the original expression.
Before accepting your proof, check: Did you cancel only factors? Did you divide by something that could be 0? Did you square or take a root and lose a sign? Does the final domain still match the original expressions?
09 / Your turn
State the original restrictions as well as the simplified result.
02 · A cot product
Simplify 4sin θ cot θ − cos θ.
Hint
Sin θ cot θ = cos θ on its domain.
Worked solution
3cos θ, with sin θ ≠ 0.
03 · Powers
Simplify cosec⁴θ sin⁶θ.
Hint
Write cosec⁴θ as 1/sin⁴θ.
Worked solution
Sin²θ, with sin θ ≠ 0. Four factors cancel; two remain.
04 · Reciprocal quotient
Simplify sec θ/cosec θ.
Hint
Divide the two reciprocal fractions.
Worked solution
(1/cos θ)/(1/sin θ) = sin θ/cos θ = tan θ, with both sin θ ≠ 0 and cos θ ≠ 0. The original denominator cosec is never 0 but must exist.
05 · Prove a difference
Prove cosec θ − sin θ = cos θ cot θ.
Hint
Use a common denominator of sin θ.
Worked solution
LHS = (1 −sin²θ)/sin θ = cos²θ/sin θ = cos θ cot θ. Require sin θ ≠ 0.
06 · A sum
Simplify sin θ cos θ(tan θ + cot θ).
Hint
Use the common-denominator identity.
Worked solution
Sin θ cos θ × 1/(sin θ cos θ) = 1, with both sin θ and cos θ nonzero.
07 · Difference of squares
Prove (sec θ −1)(sec θ +1) = tan²θ using sine and cosine.
Hint
Expand the product and write sec² as 1/cos².
Worked solution
Sec²θ −1 = (1 −cos²θ)/cos²θ = sin²θ/cos²θ = tan²θ. Require cos θ ≠ 0.
08 · Square root sign
Evaluate √(sec²θ) at θ = 4π/3.
Hint
Cos(4π/3) = −1/2.
Worked solution
Sec θ = −2, so √(sec²θ) = √4 = 2 = |sec θ|.
09 · A signed product
For π/2 < θ < π, simplify √(cosec²θ cot²θ) without an absolute-value sign.
Hint
Cosec is positive and cot is negative in this quadrant.
Worked solution
−cosec θ cot θ. The product is negative, so its absolute value is its negative.
10 · Rationalise the form
Show sin θ/(1 + cos θ) = cosec θ − cot θ on their common domain.
Hint
Multiply the left fraction by (1 −cos θ)/(1 −cos θ), after checking exclusions.
Worked solution
On sin θ ≠ 0, both 1±cos θ are nonzero. The left side becomes sin θ(1−cos θ)/(1−cos²θ) = (1−cos θ)/sin θ = cosec θ−cot θ. The left expression alone also exists at some excluded inputs, so the identity is stated on the common domain sin θ ≠ 0.
11 · Use the reciprocal pair
If cosec θ + cot θ = 3, find cosec θ − cot θ, then cosec θ and cot θ.
Hint
The two combinations have product 1. Then add and subtract.
Worked solution
The other combination is 1/3. Adding gives 2cosec θ = 10/3, so cosec θ = 5/3. Subtracting gives 2cot θ = 8/3, so cot θ = 4/3. These are consistent with sin θ = 3/5 and cos θ = 4/5.
12 · Projection length
In the projection diagram, d = 6 and θ = π/4. Find BT exactly.
Hint
Use d(sec θ −cos θ).
Worked solution
BT = 6(√2 −√2/2) = 3√2. Alternatively, 6sin(π/4)tan(π/4) = 3√2.
13 · A forbidden cancellation
A student says (sec θ + cos θ)/sec θ = 1 + cos θ by cancelling sec in the first term only. Correct the simplification.
Hint
Divide both numerator terms by sec θ.
Worked solution
The result is 1 + cos θ/sec θ = 1 + cos²θ, with cos θ ≠ 0. The second term must also be divided by the denominator.
14 · Numerical checks
An identity matches at θ = π/6 and π/4. Is that a proof? What else is needed?
Hint
A statement about every allowed angle needs a general argument.
Worked solution
No. Those are useful checks, but a false general formula may agree at selected inputs. Give valid algebraic or geometric reasoning for all angles in the stated domain.
10 / Recap
Prove the equality and retain its conditions.
- Rewrite reciprocal functions using sine and cosine.
- Use common denominators for sums and differences.
- Cancel only nonzero factors.
- Use sin²θ + cos²θ = 1 where it simplifies the expression.
- √(z²) = |z| for real z.
- Simplification does not restore excluded inputs.
- Numerical checks support a proof; they do not replace it.
Section 1 of 10 · Start with the definitions and the domain