Mathematics — CBSE Class 10 board question

Rationalise the given expression:

$$\tan A = \frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1} = \frac{(\sqrt{3}-1)^2}{3-1} = \frac{4-2\sqrt{3}}{2} = 2-\sqrt{3}$$

From Table 8.1, $\tan 15° = 2 - \sqrt{3}$, so $A = 15°$.

$$\sin A - \cos A = \sin 15° - \cos 15°$$

Using $A = 15°$: $\sin 15° = \dfrac{\sqrt{6}-\sqrt{2}}{4}$, $\cos 15° = \dfrac{\sqrt{6}+\sqrt{2}}{4}$

$$\sin A - \cos A = \frac{\sqrt{6}-\sqrt{2}}{4} - \frac{\sqrt{6}+\sqrt{2}}{4} = \frac{-2\sqrt{2}}{4} = \boxed{-\dfrac{\sqrt{2}}{2}}$$

Source: Chapter 8, Section 8.3 (Trigonometric Ratios of Specific Angles)

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• Key step: Rationalise $\tan A$ to identify the angle. $\tan 15° = 2 - \sqrt{3}$ is a standard result derivable from $\tan(45°-30°)$.
• Since $A = 15°$, use exact values: $\sin 15° = \frac{\sqrt{6}-\sqrt{2}}{4}$ and $\cos 15° = \frac{\sqrt{6}+\sqrt{2}}{4}$.
• The subtraction gives $-\frac{2\sqrt{2}}{4} = -\frac{\sqrt{2}}{2}$ (also written $-\frac{1}{\sqrt{2}}$).
• Examiners award 1 mark for correctly finding $A = 15°$ and 1 mark for the final value.