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0606/12 Feb/Mar 2025, Question 6: Trig Ratios of Any Angle

Cambridge IGCSE Additional Mathematics 0606 · Paper 12 (Non-calculator) · Non-calculator · 5 marks

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The question, in short

Given tan θ = √5/5 with 180° < θ < 360°, find cos θ, sin θ, and sec θ + cot θ in the form (a + b√c)/√5.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 2 marks

Find cos θ given tan θ = √5/5 and θ in the third or fourth quadrant.

  1. 1tan θ = 1/√5 > 0 and 180° < θ < 360°, so θ is in the third quadrant
  2. 2Right triangle: opposite 1, adjacent √5, hypotenuse √6
  3. 3cos θ is negative: cos θ = −√5/√6 = −√(5/6)

Answercos θ = −√(5/6)

Part (b) · 1 mark

Find sin θ using the given tangent and quadrant information.

  1. 1sin θ is negative in the third quadrant
  2. 2sin θ = −1/√6 = −√(1/6)

Answersin θ = −√(1/6)

Part (c) · 2 marks

Find sec θ + cot θ, giving the result as (a + b√c)/√5 with integers a, b, c.

  1. 1sec θ = 1/cos θ = −√6/√5
  2. 2cot θ = 1/tan θ = 5/√5 = √5
  3. 3sec θ + cot θ = −√6/√5 + 5/√5 = (5 − √6)/√5

Answer(5 − √6)/√5

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Also searched as: 0606/12/F/M/25 · 0606_m25_qp_12 · 0606 m25 12 · question 6.
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