A-Level Mathematics May/June 2024 Q4(a): Show that 3 tan 20+ tan(0+45°) = (tan²0+8 tan0+1)/(1-tan²0)

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  To show that $3 \tan 2\theta + \tan(\theta+45^\circ) = \frac{\tan^2\theta+8 \tan\theta+1}{1-\tan^2\theta}$, we start with the left-hand side (LHS).

  First, we express each term using standard trigonometric identities in terms of $\tan\theta$.

  For the first term, using the double angle identity $\tan 2\theta = \frac{2 \tan \theta}{1-\tan^2\theta}$: $3 \tan 2\theta = 3 \left( \frac{2 \tan \theta}{1-\tan^2\theta} \right) = \frac{6 \tan \theta}{1-\tan^2\theta}$

  For the second term, using the compound angle identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ and knowing that $\tan 45^\circ = 1$: $\tan(\theta+45^\circ) = \frac{\tan \theta + \tan 45^\circ}{1 - \tan \theta \tan 45^\circ} = \frac{\tan \theta + 1}{1 - \tan \theta}$

  Now, substitute these back into the LHS: LHS $= \frac{6 \tan \theta}{1-\tan^2\theta} + \frac{\tan \theta + 1}{1 - \tan \theta}$

  To combine these into a single fraction, we find a common denominator. We note that $1-\tan^2\theta = (1-\tan\theta)(1+\tan\theta)$.

  LHS $= \frac{6 \tan \theta}{(1-\tan\theta)(1+\tan\theta)} + \frac{(\tan \theta + 1)(1+\tan\theta)}{(1 - \tan \theta)(1+\tan\theta)}$

  LHS $= \frac{6 \tan \theta + (1 + \tan \theta)^2}{1-\tan^2\theta}$

  Now, expand the numerator: Numerator $= 6 \tan \theta + (1 + 2\tan\theta + \tan^2\theta)$ Numerator $= \tan^2\theta + 8\tan\theta + 1$

  So, the complete expression is: LHS $= \frac{\tan^2\theta + 8\tan\theta + 1}{1-\tan^2\theta}$

  This is the same as the right-hand side (RHS), as required.

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  How the marks are awarded

  • B1 — Correctly stating the identity for 3 tan 2θ in terms of tan θ. In the model answer, this is shown by writing 3 tan 2θ = (6 tan θ)/(1-tan²θ).
  • B1 — Correctly stating the identity for tan(θ+45°) using the compound angle formula and tan 45° = 1. This is shown by deriving (tan θ + 1)/(1 - tan θ).
  • M1 — Attempting to combine the two expressions into a single fraction. This involves finding the common denominator 1-tan²θ and correctly adjusting the numerators before adding them.
  • A1 — Correctly expanding and simplifying the numerator to tan²θ + 8 tanθ + 1 and presenting the final fraction to confirm the given identity, with no errors in the working.

  Common mistakes

  • Incorrectly recalling the double angle identity, e.g., using tan 2θ = (2 tan θ)/(1 + tan²θ).
  • Making an algebraic error when creating a common denominator, such as simply multiplying the denominators without recognising that 1-tan θ is a factor of 1-tan²θ.
  • Errors in expanding the numerator, for example (1 + tan θ)² = 1 + tan²θ, forgetting the middle 2 tan θ term.
  • Using the wrong compound angle formula, such as tan(A-B) instead of tan(A+B), leading to sign errors in the second term.

  Examiner tip: This question rewards fluent recall of trigonometric identities and methodical algebraic manipulation, particularly combining fractions by finding a common denominator.

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