A-Level Mathematics October/November 2024 Q4(b): Hence or otherwise solve the equation sec⁴2α-tan⁴ 2α = 2 tan² 2α sec² 2α for 0° < α < 1…

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  The equation is $sec^{4}2α - tan^{4} 2α = 2 tan^{2} 2α sec^{2} 2α$.

  The left-hand side is a difference of two squares: $(sec^{2}2α - tan^{2}2α)(sec^{2}2α + tan^{2}2α) = 2 tan^{2} 2α sec^{2} 2α$

  Using the identity sec²x - tan²x ≡ 1: $(1)(sec^{2}2α + tan^{2}2α) = 2 tan^{2} 2α sec^{2} 2α$ $sec^{2}2α + tan^{2}2α = 2 tan^{2} 2α sec^{2} 2α$

  To form an equation in terms of tan 2α, substitute $sec^{2}2α = 1 + tan^{2}2α$: $(1 + tan^{2}2α) + tan^{2}2α = 2 tan^{2} 2α (1 + tan^{2}2α)$ $1 + 2tan^{2}2α = 2tan^{2}2α + 2tan^{4}2α$

  Subtracting 2tan²2α from both sides gives: $1 = 2tan^{4}2α$ $tan^{4}2α = \frac{1}{2}$

  Now, we solve for tan 2α: $tan^{2}2α = \frac{1}{\sqrt{2}}$ $tan 2α = \pm \sqrt{\frac{1}{\sqrt{2}}} = \pm \frac{1}{\sqrt[4]{2}}$ $tan 2α \approx \pm 0.840896...$

  The range for α is 0° < α < 180°, so the range for 2α is 0° < 2α < 360°.

  Case 1: $tan 2α = +0.840896...$ The principal value is $2α = tan^{-1}(0.840896...) = 40.06...°$. The other solution in the range is 2α = 180° + 40.06...° = 220.06...°.

  Case 2: $tan 2α = -0.840896...$ The principal value is -40.06...°. The solutions in the range 0° < 2α < 360° are: 2α = 180° - 40.06...° = 139.93...°. 2α = 360° - 40.06...° = 319.93...°.

  So, the four values for 2α are 40.06...°, 139.93...°, 220.06...°, and 319.93...°.

  Dividing by 2 to find the values for α: $α = \frac{40.06...°}{2} = 20.03...° \approx 20.0°$ $α = \frac{139.93...°}{2} = 69.96...° \approx 70.0°$ $α = \frac{220.06...°}{2} = 110.03...° \approx 110.0°$ $α = \frac{319.93...°}{2} = 159.96...° \approx 160.0°$

  The solutions in the given range are α = 20.0°, 70.0°, 110.0°, 160.0°.

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  How the marks are awarded

  • M1 — Using the identities $sec^2x - tan^2x = 1$ and $sec^2x = 1 + tan^2x$ to obtain an equation in a single trigonometric function, e.g., $(1 + tan^{2}2α) + tan^{2}2α = 2 tan^{2} 2α (1 + tan^{2}2α)$.
  • M1 — Correctly simplifying the equation to $tan^{4}2α = 1/2$ and solving to find values for tan 2α, i.e., $tan 2α = \pm \sqrt{1/\sqrt{2}}$.
  • A1 — Obtaining one correct solution for α, such as α = 20.0°, from a correct value of 2α.
  • A1 — Obtaining a second correct solution for α, such as α = 70.0°, from a different correct value of 2α.
  • A1 — Obtaining the final two solutions, 110.0° and 160.0°, and having no other incorrect solutions within the specified range.

  Common mistakes

  • Forgetting the negative root when solving $tan^{2}2α = 1/\sqrt{2}$, which leads to finding only two of the four solutions.
  • Failing to correctly adjust the range. Students solve for α in 0° < α < 180° but forget to first find all solutions for 2α in the doubled range 0° < 2α < 360°, thus missing solutions.
  • Making an algebraic error when simplifying the equation, for example cancelling 2tan²2α incorrectly to get $1 = 2tan^{4}2α$ without first expanding the right-hand side.
  • Finding only the principal values for 2α (e.g., 40.06° and -40.06°) and not using the CAST diagram or tan graph properties to find all solutions in the 0° < 2α < 360° range.

  Examiner tip: Always look to simplify complex trigonometric equations by using Pythagorean identities to express them in terms of a single trigonometric function.

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