The reciprocal trig functions are sec θ = 1 / cos θ, cosec θ = 1 / sin θ and cot θ = 1 / tan θ = cos θ / sin θ. From sin²θ + cos²θ = 1 come two key identities, sec²θ = 1 + tan²θ and cosec²θ = 1 + cot²θ, which turn awkward equations into quadratics you can factorise and solve.
Reciprocal trig functions and identities
- sec θ = 1 / cos θ
- cosec θ = 1 / sin θ
- cot θ = 1 / tan θ = cos θ / sin θ
- sec²θ = 1 + tan²θ
- cosec²θ = 1 + cot²θ
The third letter tells you which function is underneath: sec goes with cos, cosec goes with sin, cot goes with tan.
What are sec, cosec and cot?
Each is the reciprocal of a familiar function. To solve an equation in one of them, flip it first and write the equivalent sin, cos or tan equation on its own line: sec θ = 2 becomes cos θ = ½, and cot θ = 3 becomes tan θ = 1/3.
The flip cot θ = k to tan θ = 1/k only works when k is not zero. If cot θ = 0, go back to cos θ / sin θ = 0, which means cos θ = 0.
How do you sketch sec, cosec and cot?
Build each graph from its base curve rather than memorising it. Sketch the base curve lightly, then apply four rules:
- Where the base curve is zero, the reciprocal has a vertical asymptote.
- Where the base curve is 1 or −1, the two curves touch, and these become turning points.
- Where the base is small, the reciprocal is large, and the reverse.
- The reciprocal always has the same sign as the base.
So each peak of y = cos θ becomes the bottom of a U on y = sec θ, and each trough becomes the top of an upside-down U. That is why sec θ and cosec θ never take values between −1 and 1: |sec θ| ≥ 1 and |cosec θ| ≥ 1.
Where do sec²θ = 1 + tan²θ and cosec²θ = 1 + cot²θ come from?
Both come from sin²θ + cos²θ = 1, so there is nothing new to memorise. Divide every term by cos²θ to get tan²θ + 1 = sec²θ. Divide every term by sin²θ to get 1 + cot²θ = cosec²θ.
To choose between them, match the functions in the question: sec pairs with tan, and cosec pairs with cot.
Equations that become quadratics
An equation mixing sec² with tan, or cosec² with cot, becomes a quadratic once you substitute the identity. Factorise, then flip each root back to sin, cos or tan before solving.
Watch for impossible roots. Because |sec θ| ≥ 1, a root such as sec θ = ½ must be rejected, since it would need cos θ = 2.
Prove, then hence solve
The classic Pure 3 question proves an identity in part (a) and says “hence solve” in part (b). Part (b) usually carries more marks and depends on using the result from (a). When the equation has a common factor such as cot x, factorise rather than divide, and treat the cot x = 0 branch carefully: it means cos x = 0, not tan x = 1/0.
Worked examples
Example 1. Find the exact value of sec(5π/6), giving your answer in the form a√3 / 3.
- Flip to cosine: sec(5π/6) = 1 / cos(5π/6).
- 5π/6 = 150°, and cos 150° = −√3/2.
- Reciprocal: 1 / (−√3/2) = −2/√3.
- Rationalise: −2/√3 × √3/√3 = −2√3/3, so a = −2.
Answer: sec(5π/6) = −2√3/3.
Marks: M1 for the reciprocal of cos, A1 for the exact surd.
Example 2. Given that tan θ = 3/4 and θ is acute, find the exact value of sec θ.
- Identity: sec²θ = 1 + tan²θ = 1 + 9/16 = 25/16.
- Square root, keeping ±: sec θ = ±5/4.
- θ is acute, so cos θ > 0 and sec θ > 0.
Answer: sec θ = 5/4.
Marks: M1 for the identity, A1 for the sign and value.
Example 3. Solve sec²θ + tan θ = 3 for 0° ≤ θ < 360°, to 1 d.p. where appropriate.
- Substitute: (1 + tan²θ) + tan θ = 3.
- Collect: tan²θ + tan θ − 2 = 0.
- Factorise: (tan θ − 1)(tan θ + 2) = 0, so tan θ = 1 or tan θ = −2.
- tan θ = 1: θ = 45°, 225°.
- tan θ = −2: the base angle is 63.4°, so θ = 116.6°, 296.6°.
Answer: θ = 45°, 116.6°, 225°, 296.6°.
Marks: M1 for the identity, M1 for factorising, A1 for all four.
Example 4. (a) Prove that cosec x − sin x ≡ cos x cot x. (b) Hence solve cosec x − sin x = ½ cot x for 0° < x < 360°.
- (a) cosec x − sin x = 1/sin x − sin x = (1 − sin²x) / sin x = cos²x / sin x = cos x × (cos x / sin x) = cos x cot x.
- (b) Using part (a): cos x cot x = ½ cot x.
- Factorise, without dividing by cot x: cot x(cos x − ½) = 0.
- cot x = 0 means cos x / sin x = 0, so cos x = 0: x = 90°, 270°.
- cos x = ½: x = 60°, 300°.
Answer: x = 60°, 90°, 270°, 300°.
Marks: M1 for the proof, M1 for factorising, A1 for the cot x = 0 branch, A1 for all four.
Common mistakes
- Pairing sec with sin. sec θ = 1 / cos θ and cosec θ = 1 / sin θ. The third letter tells you: se(c) goes with cos, co(s)ec goes with sin, co(t) goes with tan.
- Dividing by cot x or tan x. Factorise instead. Dividing loses every solution where that function is zero.
- Flipping cot x = 0 to tan x = 1/0. cot x = 0 means cos x = 0, which gives x = 90° and 270°.
- Keeping impossible roots. sec θ and cosec θ are never between −1 and 1, so a root like sec θ = ½ must be rejected.
- Dropping the ± after a square root. sec²θ = 25/16 gives sec θ = ±5/4. Use the quadrant to choose the sign.
How sec, cosec and cot are examined
In Edexcel International A-Level these are Pure 3 content (paper WMA13), in the trigonometric functions chapter. In Edexcel A-Level (9MA0) they are Year 2 Pure. Typical questions:
- sketch y = sec θ, cosec θ or cot θ, or a transformation, showing asymptotes and turning points
- find an exact value such as sec(5π/6) or cosec 240°
- prove an identity using sec² = 1 + tan² or cosec² = 1 + cot²
- solve an equation that becomes a quadratic, often as the “hence” part after a proof
Work through the Pure 3 trigonometric functions exam questions with mark schemes, then the trigonometric functions worksheet. For unlimited mixed practice, use the trigonometry revision engine.
Related courses: inverse trig functions and further trig equations, which combine these identities with double angles.
Frequently asked questions
What is the difference between sec and cos⁻¹?
sec θ = 1 / cos θ is a reciprocal. cos⁻¹ is the inverse function, arccos, which gives an angle. They are completely different.
Do I need to memorise sec²θ = 1 + tan²θ?
You can derive it in one line by dividing sin²θ + cos²θ = 1 by cos²θ, so learn that step and you will never need to look it up.
What is the range of sec θ?
sec θ ≤ −1 or sec θ ≥ 1. It never takes values between −1 and 1, because cos θ never exceeds 1 in size.