From the textbook: $\dfrac{2\tan 30°}{1+\tan^2 30°} = \sin 60°$ and $\dfrac{2\tan 30°}{1-\tan^2 30°} = \tan 60°$.
So the equation becomes $x\sin 60° = y\tan 60°$, i.e., $x \cdot \dfrac{\sqrt{3}}{2} = y \cdot \sqrt{3}$.
$$\frac{x}{y} = \frac{\sqrt{3}}{\frac{\sqrt{3}}{2}} = 2 \implies x:y = 2:1$$
Answer: (C) 2 : 1
Source: Chapter 8, Exercise 8.2, Q2(i) and Q2(iv)
The key is recognising the double-angle identities hidden in the fractions: $\dfrac{2\tan\theta}{1+\tan^2\theta} = \sin 2\theta$ and $\dfrac{2\tan\theta}{1-\tan^2\theta} = \tan 2\theta$. For θ = 30°, these give sin 60° and tan 60° respectively. The textbook Exercise 8.2 Q2 directly establishes these values, so substituting and simplifying the ratio is straightforward. Examiners expect you to recall these standard results.