Question

The value of the expression $\sqrt{3}cosec20^{\circ }-\sec 20^{\circ }$ 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)

$\sqrt{3}cosec20^{\circ }-\sec 20^{\circ }=\frac{\sqrt{3}}{\sin 20^{\circ }}-\frac{1}{\cos 20^{\circ }}$
$=\frac{\sqrt{3}\cos 20^{\circ }-\sin 20^{\circ }}{\sin 20^{\circ }\cos 20^{\circ }}$
$=\frac{4\left(\frac{\sqrt{3}}{2}\cos 20^{\circ }-\frac{1}{2}\sin 20^{\circ }\right)}{2\sin 20^{\circ }\cos 20^{\circ }}$
$=\frac{4\sin \left(60^{\circ }-20^{\circ }\right)}{\sin \left(2\times 20^{\circ }\right)}=4$