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A-LevelMathematicsTrigonometry9709/31 · Oct/Nov 2024 · Q48 Marks

Hence or otherwise solve the equation sec⁴2α-tan⁴ 2α = 2 tan² 2α sec² 2α for 0° < α < 180°.

  1. (b)Hence or otherwise solve the equation sec⁴2α-tan⁴ 2α = 2 tan² 2α sec² 2α for 0° < α < 180°.[5]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2024). The mark or mark type is shown in brackets.

  1. [M1] Form an equation in tan 2α. 1+2tan² 2α = 2tan² 2α(1+tan² 2α) Or multiply through by cos⁴ 2α to form an equation in sin 2α or cos 2α cos⁴ 2α + 2sin² 2α cos² 2α = 2sin⁴ 2α ⇒ sin⁴ 2α +2sin² 2α-1=0 or cos⁴ 2α-4cos² 2α +2 = 0
  2. [M1] Solve for tan 2α or equivalent tan 2α = ±(1/√2)
  3. [A1] Obtain one correct solution for α, e.g. 20.0(30..)°
  4. [A1] Obtain a second correct value for α, e.g. 70.0° (69.9698..°)
  5. [A1] Obtain solutions 110° (110.0) and 160° (160.0) for α, and no others in range

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About this question

Topic
Trigonometry, A-Level Mathematics (syllabus 9709). Structured question, 8 marks.
Source
A-Level Mathematics, Oct/Nov 2024 Paper 3 Variant 1, question 4.

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