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 $sec⁴2α - tan⁴ 2α = 2 tan² 2α sec² 2α$.

  The left-hand side is a difference of two squares: $(sec²2α - tan²2α)(sec²2α + tan²2α) = 2 tan² 2α sec² 2α$

  Using the identity $sec²x - tan²x ≡ 1$: $(1)(sec²2α + tan²2α) = 2 tan² 2α sec² 2α$ $sec²2α + tan²2α = 2 tan² 2α sec² 2α$

  To form an equation in terms of $tan 2α$, substitute $sec²2α = 1 + tan²2α$: $(1 + tan²2α) + tan²2α = 2 tan² 2α (1 + tan²2α)$ $1 + 2tan²2α = 2tan²2α + 2tan⁴2α$

  Subtracting $2tan²2α$ from both sides gives: $1 = 2tan⁴2α$ $tan⁴2α = \frac{1}{2}$

  Now, we solve for $tan 2α$: $tan²2α = \frac{1}{\sqrt{2}}$ $tan 2α = ±\sqrt{\frac{1}{\sqrt{2}}} = ±\frac{1}{\sqrt[4]{2}}$ $tan 2α ≈ ±0.840896...$

  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 $α$: $α = \frac{40.06...°}{2} = 20.03...° ≈ 20.0°$ $α = \frac{139.93...°}{2} = 69.96...° ≈ 70.0°$ $α = \frac{220.06...°}{2} = 110.03...° ≈ 110.0°$ $α = \frac{319.93...°}{2} = 159.96...° ≈ 160.0°$

  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., $(1 + tan²2α) + tan²2α = 2 tan² 2α (1 + tan²2α)$.
  • M1 — Correctly simplifying the equation to $tan⁴2α = 1/2$ and solving to find values for $tan 2α$, i.e., $tan 2α = ±\sqrt{1/\sqrt{2}}$.
  • 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 $tan²2α = 1/\sqrt{2}$, 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 $1 = 2tan⁴2α$ 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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