Generated by claude-sonnet-4-6 · 2026-06-15 10:23 · grounding rag
Model Answer
Substituting standard values: $\tan 30° = \dfrac{1}{\sqrt{3}}$, $\sec 60° = 2$, $\tan 45° = 1$, $\sin 60° = \dfrac{\sqrt{3}}{2}$
$$\sin^2 60° = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$$
$$\frac{2 \times \dfrac{1}{\sqrt{3}} \times 2 \times 1}{1 - \dfrac{3}{4}} = \frac{\dfrac{4}{\sqrt{3}}}{\dfrac{1}{4}} = \frac{4}{\sqrt{3}} \times 4 = \frac{16}{\sqrt{3}} = \frac{16\sqrt{3}}{3}$$
$$\boxed{\dfrac{16\sqrt{3}}{3}}$$
Source: Chapter 8, Section 8.3 (Table 8.1)
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Explanation
- Examiners expect you to clearly substitute each value from the standard table before simplifying.
- The denominator uses $1 - \sin^2 60° = 1 - \frac{3}{4} = \frac{1}{4}$ — a common error is computing $\cos^2 60°$ instead, but here it is $1 - \sin^2 60°$, which equals $\cos^2 60° = \frac{1}{4}$ by identity — same result, just be careful.
- Rationalise the final answer: $\frac{16}{\sqrt{3}} = \frac{16\sqrt{3}}{3}$.