First slide

Compound angles

Question

$\tan 81^{\circ} - \tan 63^{\circ} - \tan 27^{\circ} + \tan 9^{\circ}$ equals

Moderate

A

6
B

0
C

2
D

4

Solution

$\begin{array}{l} \tan 81^{\circ} - \tan 63^{\circ} - \tan 27^{\circ} + \tan 9^{\circ} = [\tan 81^{\circ} + \tan 9^{\circ}) - [\tan (63^{\circ}) + \tan (27^{\circ})] \\ = [\frac{\sin 81^{\circ}}{\cos 81^{\circ}} + \frac{\sin 9^{\circ}}{\cos 9^{\circ}}] - [\frac{\sin 63^{\circ}}{\cos 63^{\circ}} + \frac{\sin 27^{\circ}}{\cos 27^{\circ}}] \\ = [\frac{\sin 81^{\circ} \cos 9^{\circ} + \sin 9^{\circ} \cos 81^{\circ}}{\cos 9^{\circ} \cos 81^{\circ}}] - [\frac{\sin 63^{\circ} \cos 27^{\circ} + \sin 27^{\circ} \cos 63^{\circ}}{\cos 27^{\circ} \cos 63^{\circ}}] \\ = \frac{[\sin (81^{\circ} + 9^{\circ})]}{\cos 9^{\circ} \cos (90^{\circ} - 9^{\circ})} - \frac{[\sin (63^{\circ} + 27^{\circ})]}{\cos 27^{\circ} \cos (90^{\circ} - 27^{\circ})} \\ = \frac{\sin 90^{\circ}}{\cos 9^{\circ} \sin 9^{\circ}} - \frac{\sin 90^{\circ}}{\cos 27^{\circ} \sin 27^{\circ}} \\ = \frac{1}{\sin 9^{\circ} \cos 9^{\circ}} \times \frac{2}{2} - \frac{1}{\sin 27^{\circ} \cos 27^{\circ}} \times \frac{2}{2} \\ = \frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}} \\ = 2 [\frac{\sin 54^{\circ} - \sin 18^{\circ}}{\sin 18^{\circ} \sin 54^{\circ}}] = 2 [\frac{2 \cos 36^{\circ} \sin 18^{\circ}}{\sin 18^{\circ} \cos 36^{\circ}}] = 4 \end{array}$

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