Mathematics — Standard (041) — CBSE Class 10 board question

Substituting standard values: $\cos 60° = \dfrac{1}{2}$, $\sec 30° = \dfrac{2}{\sqrt{3}}$, $\tan 45° = 1$, $\sin 30° = \dfrac{1}{2}$, $\sin 60° = \dfrac{\sqrt{3}}{2}$

Numerator:
$$5\left(\frac{1}{2}\right)^2 + 4\left(\frac{2}{\sqrt{3}}\right)^2 - (1)^2 = 5 \times \frac{1}{4} + 4 \times \frac{4}{3} - 1 = \frac{5}{4} + \frac{16}{3} - 1 = \frac{15 + 64 - 12}{12} = \frac{67}{12}$$

Denominator:
$$\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1$$

$$\therefore \quad \frac{67/12}{1} = \boxed{\dfrac{67}{12}}$$

Source: Introduction to Trigonometry, Section 8.3

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• The denominator $\sin^2 30° + \sin^2 60° = 1$ (not $\sin^2 30° + \cos^2 30°$ as in the textbook Exercise 8.2(v) — note the question has $\sin^2 60°$ in the denominator, giving the same result since $\sin 60° = \cos 30°$).
• The key step is careful substitution from Table 8.1 and finding the LCM (12) for the numerator.
• Examiners expect all substitution steps shown clearly to earn both marks.