Interpret reciprocal trigonometric functions and their domains
Use secant, cosecant and cotangent as reciprocal ratios, keeping their definitions separate from principal inverse angles.
Exam scope and assessment notes
Pure Mathematics 2 (Paper 2, AS only) and Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027. Paper 1 knowledge is assumed. The AS route using Papers 1 + 2 cannot be carried forward to A Level. Shared sections 2.3 and 3.3 cover all six trigonometric functions, reciprocal squared identities, compound and double angles, and amplitude-phase forms used in simplification, exact evaluation and equations. State angle units, keep original domains and find every solution in the supplied interval. Written practice is self-checked.
Before you begin
- Use exact sine, cosine and tangent values and quadrant signs.
- Distinguish an inverse function from a reciprocal and avoid division by zero.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does cosec 30° equal?
What you will learn
- Evaluate sec, cosec and cot for exact, negative and multi-turn angles.
- Identify the original denominator restrictions and distinguish reciprocal values from inverse angles.
Three reciprocal functions extend the familiar ratios
Secant is sec θ = 1/cos θ where cos θ ≠ 0. Cosecant is cosec θ = 1/sin θ where sin θ ≠ 0; the notation csc means the same function. Cotangent is cot θ = cos θ/sin θ where sin θ ≠ 0. It also equals 1/tan θ when tangent is defined and non-zero, but that reciprocal expression has a narrower domain than the cosine-over-sine definition. At 90°, cot θ = 0/1 = 0 even though tangent is undefined.
These are reciprocal numerical ratios, not principal inverse angles. At θ = 30°, cosec θ = 2, while sin⁻¹(1/2) = 30°. Sine's inverse accepts a ratio and returns a selected angle; cosecant accepts an angle and returns a reciprocal ratio. Use special-angle values first, then invert a non-zero value exactly rather than round it prematurely.
Quadrants and whole turns still determine the sign
A reciprocal keeps the sign of its non-zero denominator. Thus sec 150° = −2/√3 = −2√3/3 and cosec(−π/6) = −2. Cotangent takes the sign of cosine divided by sine. At 5π/4 both are negative, so cot(5π/4) = 1. Secant and cosecant repeat after 2π radians or 360°, while cotangent repeats after π radians or 180°; reducing a multi-turn input must respect the chosen function's period. For example sec(7π/3) = sec(π/3) = 2, while cot(9π/4) = cot(π/4) = 1 after subtracting two cotangent periods.
If sin θ = 3/5 and θ lies in quadrant II, then cos θ = −4/5 from the squared identity and quadrant sign. The six ratios are sin = 3/5, cos = −4/5, tan = −3/4, sec = −5/4, cosec = 5/3 and cot = −4/3. The interval or quadrant matters when recovering the missing coordinate: the squared identity alone does not choose its sign.
Pause and explain
Which statement about cot 90° is correct?
Worked example
Find sec(5π/3), cosec(7π/6) and cot(3π/2) exactly. Is sec(π/2) defined?
Show the worked solution
- cos(5π/3) = 1/2, so its reciprocal sec(5π/3) is 2.
- sin(7π/6) = −1/2, so cosec(7π/6) = −2; at 3π/2 cosine is zero and sine is −1, so cot is 0.
- At π/2 cosine is zero, making sec(π/2) undefined; do not confuse this with the valid zero of cotangent.
Answer sec(5π/3) = 2; cosec(7π/6) = −2; cot(3π/2) = 0; sec(π/2) is undefined.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find sec 120°, cosec 120° and cot 120° exactly.
Give me a hint
Start with the quadrant-II sine and cosine values, then use the defining ratios.
Compare my reasoning
- At 120°, sin = √3/2 and cos = −1/2.
- Their reciprocals give sec = −2 and cosec = 2/√3 = 2√3/3.
- The cosine-over-sine quotient gives cot = −1/√3 = −√3/3.
sec 120° = −2; cosec 120° = 2√3/3; cot 120° = −√3/3.
Look for these in your work
- Retain the quadrant signs in the original ratios.
- Evaluate and, where helpful, rationalise each reciprocal exactly.
Evaluate sec(−π/3), cosec(5π/6) and cot(7π/4) exactly.
Give me a hint
Use the cosine evenness, quadrant-II sine and quadrant-IV quotient separately.
Compare my reasoning
- cos(−π/3) = 1/2, so sec(−π/3) = 2.
- sin(5π/6) = 1/2, so cosec(5π/6) = 2.
- At 7π/4 cosine is √2/2 and sine is −√2/2, giving cot = −1.
2, 2 and −1 respectively.
Look for these in your work
- Identify signs before taking reciprocals.
- Use cotangent's defining quotient for the supplied angle.
A supplied angle θ is in quadrant II and has sin θ = 3/5. Find the other five trigonometric ratios exactly.
Give me a hint
Use the squared identity to find cosine's magnitude, then let the quadrant choose its sign.
Compare my reasoning
- cos² θ = 1 − 9/25 = 16/25; quadrant II gives cos θ = −4/5.
- Then tan θ = (3/5)/(−4/5) = −3/4 and sec θ = −5/4.
- The remaining reciprocals are cosec θ = 5/3 and cot θ = −4/3; all original denominators are non-zero.
cos θ = −4/5; tan θ = −3/4; sec θ = −5/4; cosec θ = 5/3; cot θ = −4/3.
Look for these in your work
- Use the quadrant to resolve the square-root sign.
- Check that each defining denominator is non-zero.
Common mistakes
- A reciprocal ratio is different from a principal inverse angle.
- Use cot θ = cos θ/sin θ to retain its full domain; 1/tan θ cannot be used at every valid cotangent input.
- Keep the quadrant sign when taking a reciprocal or recovering a missing ratio.
What can you explain now?
Evaluate cot 90° and explain why 1/tan 90° is not a valid calculation.
Compare with the explanation
cot 90° = 0, from cos 90°/sin 90° = 0/1.
Tangent is undefined at 90°; its reciprocal expression cannot be evaluated there even though cotangent's defining quotient can.
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Make it stick
Define secant, cosecant and cotangent with their denominator restrictions. Give an angle where cotangent is valid but its 1/tan expression cannot be evaluated.
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