(c) tan 60°
$\tan 30° = \dfrac{1}{\sqrt{3}}$, so $\dfrac{2\tan 30°}{1-\tan^2 30°} = \dfrac{2 \times \frac{1}{\sqrt{3}}}{1 - \frac{1}{3}} = \dfrac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} = \dfrac{2}{\sqrt{3}} \times \dfrac{3}{2} = \sqrt{3} = \tan 60°$
Source: Chapter 8, Exercise 8.2 Q2(iv)
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This uses the double-angle formula for tan: $\tan 2\theta = \dfrac{2\tan\theta}{1-\tan^2\theta}$. Here $\theta = 30°$, so the expression equals $\tan 60° = \sqrt{3}$. Examiners expect you to substitute the value of $\tan 30°$ and simplify step-by-step. Note: Q2(i) in the textbook uses $1 + \tan^2 30°$ (giving sin 60°), while Q2(iv) uses $1 - \tan^2 30°$ (giving tan 60°) — don't mix them up.