Question

The value of the expression $\sqrt{3} cosec 20^{\circ} -\sec 20^{\circ}$ is equal to

• A. $\frac{4 \, \sin \, 20^{\circ}}{\sin \, 40^{\circ}}$
• B. $\frac{2\, \sin \, 20^{\circ}}{\sin \, 40^{\circ}}$
• C. 4
• D. 2

Answer 1

Answer: (C)

$\sqrt{3} cosec 20^{\circ} -\sec 20^{\circ} = \frac{\sqrt{3}}{\sin20^{\circ}} - \frac{1}{\cos 20^{\circ}} $
$ = \frac{\sqrt{3} \cos 20^{\circ} -\sin 20^{\circ}}{ \sin20^{\circ} \cos 20^{\circ}} $
$ = \frac{4 \left(\frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin20^{\circ}\right)}{2\sin20^{\circ}\cos 20^{\circ}} $
$ = \frac{4\sin\left(60^{\circ} -20^{\circ}\right)}{\sin\left(2 \times20^{\circ}\right)} = 4 $