Solve equations with secant, cosecant and cotangent
Rewrite reciprocal-function equations using their definitions and Pythagorean identities, then solve and check the original domain.
Cambridge IGCSE Additional Mathematics 0606, 2028–2030 syllabus, Version 1. Paper 1 is non-calculator; Paper 2 requires a scientific calculator. These original lessons provide self-checked topic practice, not a Cambridge grade or a timed paper. Full-paper assessment and human specialist review are still outstanding.
Before you begin
- Cambridge assumes knowledge of IGCSE Mathematics or an equivalent course. Check the domain before using an inverse or logarithm, and retain exact values where the question asks for them.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which ratio defines cot θ?
What you will learn
- Solve equations involving sec, cosec or cot on a specified domain
- Use the supplied identities to reduce equations to a familiar function
- Check every candidate in the original equation and preserve exclusions
Rewrite reciprocal functions carefully
Use sec θ = 1/cos θ, cosec θ = 1/sin θ and cot θ = 1/tan θ = cos θ/sin θ. These forms expose undefined values and let you use the sine, cosine and tangent equations already known.
The supplied identities can reduce squared terms: sec²θ = 1 + tan²θ and cosec²θ = 1 + cot²θ. Keep track of which values are excluded before multiplying or dividing.
Solve, filter and substitute back
For 2sec²θ + tan θ − 3 = 0, substitute sec²θ = 1 + tan²θ and set u = tan θ. The equation becomes 2u² + u − 1 = 0, so u = 1/2 or u = −1.
Solve each tangent equation on the full stated interval. Then check that cosine is not zero and substitute a representative value into the original equation. Squaring or multiplying by a trigonometric function can introduce candidates that need this final check.
For a sum such as 5 sin 3θ + 2 cos 3θ = 0, check cos 3θ = 0 separately; it would make the left side ±5, not zero. You may then divide by cos 3θ and solve tan 3θ = −2/5, remembering to transform the domain for 3θ.
Pause and explain
If sec θ = 2, what is cos θ?
Worked example
Solve 2sec²θ + tan θ − 3 = 0 for 0 ≤ θ < 2π. Give solutions to 3 decimal places where needed.
Show the worked solution
- Replace sec²θ with 1 + tan²θ: 2tan²θ + tan θ − 1 = 0.
- Factor to (2tan θ − 1)(tan θ + 1) = 0, so tan θ = 1/2 or tan θ = −1.
- Let α = arctan(1/2). The first equation gives α and π + α; the second gives 3π/4 and 7π/4. All have cos θ ≠ 0.
Answer θ = α, π + α, 3π/4 or 7π/4, where α = arctan(1/2) ≈ 0.464 radians.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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.
Solve 2 cosec θ = 4 for 0 ≤ θ < 2π.
Give me a hint
Rewrite cosecant as 1/sin θ and check that the resulting sine value is possible.
Compare my reasoning
- 2/sin θ = 4, so sin θ = 1/2.
- Sine is positive in Quadrants I and II with reference angle π/6.
- The solutions are π/6 and 5π/6, where sine is nonzero.
θ = π/6 or 5π/6.
Look for these in your work
- Reciprocal equation rearranged
- Both sine solutions included
- Original denominator checked
Solve 4 cot θ = tan θ for 0 ≤ θ < 2π.
Give me a hint
Use cot θ = 1/tan θ, but first state that sine and cosine must be nonzero.
Compare my reasoning
- For defined cot and tan, 4/tan θ = tan θ, so tan²θ = 4.
- Thus tan θ = 2 or tan θ = −2; define α = arctan 2.
- The solutions are α, π − α, π + α and 2π − α.
θ = α, π − α, π + α or 2π − α, where α = arctan 2; none make sine or cosine zero.
Look for these in your work
- Domain exclusions stated
- Both tangent signs solved
- Four interval solutions listed
Solve sec²θ = 4 for 0 ≤ θ < 2π.
Give me a hint
Use sec²θ = 1 + tan²θ or take sec θ = ±2, then check every quadrant solution.
Compare my reasoning
- sec²θ = 4 means sec θ = 2 or −2, so cos θ = 1/2 or −1/2.
- cos θ = 1/2 at π/3 and 5π/3; cos θ = −1/2 at 2π/3 and 4π/3.
- All four have nonzero cosine and satisfy the original squared equation.
θ = π/3, 2π/3, 4π/3 or 5π/3.
Look for these in your work
- Both secant signs considered
- All cosine quadrants included
- Original equation checked
Common mistakes
- Replacing sec²θ with tan²θ instead of 1 + tan²θ
- Taking a reciprocal without retaining where the original function is undefined
- Losing solutions when dividing by a trigonometric expression that may be zero
- Failing to check candidates in the original equation
What can you explain now?
Which identity turns cosec²θ − 1 into an expression in cot θ?
Compare with the explanation
cot²θ
Cosec²θ = 1 + cot²θ, so subtracting 1 leaves cot²θ.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
When solving a reciprocal-function equation, which definitions, identities and domain checks help prevent lost or extra solutions?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.