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WAEC 2018 · Paper 2 · Q1

  1. (a)

    If tan⁡θ+tan⁡30∘1−tan⁡θtan⁡30∘+1=0\dfrac{\tan\theta + \tan30^\circ}{1 - \tan\theta\tan30^\circ} + 1 = 0, find tan⁡θ\tan\theta, leaving the answer in surd form.

Worked solution (try it first)
  1. The left side is the formula for tan⁡(θ+30∘)\tan(\theta + 30^\circ), so tan⁡(θ+30∘)=−1\tan(\theta + 30^\circ) = -1.
  2. We can also solve directly: put tan⁡30∘=13\tan 30^\circ = \frac{1}{\sqrt3} and write t=tan⁡θt = \tan\theta.
  3. Multiply the top and bottom of the fraction by 3\sqrt3: 3 t+13−t=−1\dfrac{\sqrt3\,t + 1}{\sqrt3 - t} = -1.
  4. Multiply both sides by 3−t\sqrt3 - t: 3 t+1=t−3\sqrt3\,t + 1 = t - \sqrt3.
  5. Collect the tt terms: 3 t−t=−3−1\sqrt3\,t - t = -\sqrt3 - 1, so t(3−1)=−(3+1)t(\sqrt3 - 1) = -(\sqrt3 + 1) and t=−3+13−1t = -\dfrac{\sqrt3 + 1}{\sqrt3 - 1}.
  6. Rationalise by multiplying top and bottom by 3+1\sqrt3 + 1: t=−(3+1)23−1t = -\dfrac{(\sqrt3 + 1)^2}{3 - 1}
    =−4+232= -\dfrac{4 + 2\sqrt3}{2}.
  7. So tan⁡θ=−2−3\tan\theta = -2 - \sqrt3.

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