Use squared reciprocal identities to solve trigonometric equations
Derive the secant and cosecant identities, reduce equations to one ratio and recover every permitted angle.
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 sin² θ + cos² θ = 1 and solve quadratics.
- Find all tangent solutions in a degree interval and check reciprocal denominators.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which identity connects secant and tangent?
What you will learn
- Use sec² θ = 1 + tan² θ and cosec² θ = 1 + cot² θ on their original domains.
- Solve quadratic ratio equations, retaining valid roots and all angles in the supplied interval.
Division of the squared identity produces two new identities
Divide sin² θ + cos² θ = 1 by cos² θ where cosine is non-zero. This gives tan² θ + 1 = sec² θ. Divide instead by sin² θ where sine is non-zero to obtain 1 + cot² θ = cosec² θ. The identities inherit those denominator restrictions. For example sec² θ − tan² θ = 1 where cos θ ≠ 0; neither original term can be evaluated at a cosine zero even though the simplified constant can.
Choose an identity that puts an equation into one ratio. For 2 sec² θ − tan θ = 5, set u = tan θ and replace sec² θ by 1 + u². The equation becomes 2u² − u − 3 = (2u − 3)(u + 1) = 0, giving u = 3/2 or −1. Each root can supply two angles in a full turn. No sine-or-cosine bounded-range filter applies to a finite tangent value, but cosine-zero inputs remain excluded.
Keep the original domain when multiplying by a ratio
For tan θ + cot θ = 4, both sine and cosine must be non-zero. With u = tan θ ≠ 0, the equation is u + 1/u = 4, so u² − 4u + 1 = 0 and u = 2 ± √3. The corresponding exact angles on 0° ≤ θ < 360° are 15°, 75°, 195° and 255°. Multiplying by u is valid only after keeping u ≠ 0; any zero candidate in another equation would need rejection.
If a quadratic uses cot θ instead, keep sin θ ≠ 0. For cosec² θ − 3 cot θ − 5 = 0, putting v = cot θ gives v² − 3v − 4 = 0, so v = 4 or −1. These non-zero roots can be converted to tangent values 1/4 and −1. A zero cotangent root, if present, must instead be solved directly at cosine-zero angles, rather than invert zero or discard a valid input.
Pause and explain
The equation 2u² − u − 3 = 0 comes from u = tan θ. Which roots must be used?
Worked example
Solve 2 sec² θ − tan θ = 5 for 0° ≤ θ < 360°. Give non-exact angles to one decimal place.
Show the worked solution
- The original domain excludes cos θ = 0. Replace sec² θ by 1 + tan² θ and factor (2 tan θ − 3)(tan θ + 1) = 0.
- For tan θ = 3/2, the two angles are arctan(3/2) ≈ 56.3° and 236.3°.
- For tan θ = −1, the angles are 135° and 315°. All four have non-zero cosine and belong to the half-open interval.
Answer θ ≈ 56.3°, 135°, 236.3° or 315°; exact 135° and 315° retained.
Try a topic test
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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.
Derive sec² θ − tan² θ = 1 and verify it at θ = 60°, stating the restriction.
Give me a hint
Divide the squared sine-cosine identity by cos² θ before rearranging.
Compare my reasoning
- For cos θ ≠ 0, division gives tan² θ + 1 = sec² θ.
- Rearrange to sec² θ − tan² θ = 1 on that domain.
- At 60°, sec = 2 and tan = √3, so the difference of squares is 4 − 3 = 1.
sec² θ − tan² θ = 1 for cos θ ≠ 0; at 60° it gives 4 − 3 = 1.
Look for these in your work
- Use a general identity argument, not only a numerical check.
- Retain the denominator restriction after simplifying.
Solve cosec² θ − 3 cot θ − 5 = 0 for 0° ≤ θ < 360°, giving non-exact angles to one decimal place.
Give me a hint
Use cosec² θ = 1 + cot² θ and solve for the cotangent value first.
Compare my reasoning
- With sin θ ≠ 0, the quadratic is cot² θ − 3 cot θ − 4 = (cot θ − 4)(cot θ + 1) = 0.
- Cotangent 4 gives tangent 1/4 and angles approximately 14.0° and 194.0°; cotangent −1 gives tangent −1 and exact angles 135° and 315°.
- All four original sine denominators are non-zero and all inputs belong to the supplied interval.
θ ≈ 14.0°, 135°, 194.0° or 315°.
Look for these in your work
- Choose the cosecant-cotangent identity to use a single variable.
- Recover every angle and check the original sine domain.
Solve tan θ + cot θ = 4 for 0° ≤ θ < 360°, and state the restrictions needed before multiplying by tan θ.
Give me a hint
Both original ratios must be defined; use a non-zero tangent variable to form a quadratic.
Compare my reasoning
- Require sin θ ≠ 0 and cos θ ≠ 0, so u = tan θ is defined and non-zero.
- The equation u + 1/u = 4 becomes u² − 4u + 1 = 0, giving u = 2 − √3 or 2 + √3.
- These are tan 15° and tan 75°; include their π-shifted degree values 195° and 255°. Each angle satisfies both original denominators and the equation.
θ = 15°, 75°, 195° or 255°; require sin θ ≠ 0 and cos θ ≠ 0.
Look for these in your work
- Keep the non-zero ratio condition when clearing its denominator.
- Retain both quadratic roots and all associated interval angles.
Common mistakes
- Derive each identity on its non-zero denominator domain.
- A quadratic ratio can have two roots, each giving several interval angles.
- Cotangent zero is valid where sine is non-zero; do not replace it by an invalid reciprocal of zero.
What can you explain now?
Simplify sec² θ − tan² θ and state its original-domain restriction.
Compare with the explanation
1, with cos θ ≠ 0.
Subtract tan² θ from the secant identity, retaining the domain of both original terms.
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Make it stick
Derive the two squared reciprocal identities. Explain how the original denominator and a substituted ratio's domain control which quadratic candidates can be kept.
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.