First slide

Multiple and sub- multiple Angles

Question

$\sqrt{3} cosec 20^{\circ} - \sec 20^{\circ}$ is equal to

Easy

A

1
B

2
C

3
D

4

Solution

$\begin{array}{l} \sqrt{3} cosec 20^{\circ} - \sec 20^{\circ} \\ = \frac{\sqrt{3}}{\sin 20^{\circ}} - \frac{1}{\cos 20^{\circ}} = \frac{\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ}}{\sin 20^{\circ} \cdot \cos 20^{\circ}} \\ = \frac{4 (\frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ})}{2 \sin 20^{\circ} \cdot \cos 20^{\circ}} \\ = \frac{4 (\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ})}{\sin 40^{\circ}} \\ = \frac{4 \sin 40^{\circ}}{\sin 40^{\circ}} = 4 \end{array}$

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