Question

$\frac{\sqrt{3}}{\sin (20^{\circ })}-\frac{1}{\cos (20^{\circ })}=$

• A. 1
• B. $\frac{1}{\sqrt{2}}$
• C. 2
• D. 4
• E. 0

Answer 1

Answer: (D)

$\frac{\sqrt{3}}{\sin \left(20^{\circ }\right)}-\frac{1}{\cos \left(20^{\circ }\right)}$
$=\frac{\sqrt{3}\cos \left(20^{\circ }\right)-\sin \left(20^{\circ }\right)}{\sin \left(20^{\circ }\right)\cos \left(20^{\circ }\right)}$
$=\frac{4\left[\frac{\sqrt{3}}{2}\cos \left(20^{\circ }\right)-\frac{\sin \left(20^{\circ }\right)}{2}\right]}{2\sin \left(20^{\circ }\right)\cos \left(20^{\circ }\right)}$
$[\because 2\sin A\cos A=\sin 2A]$
$=\frac{4\left(\sin 60^{\circ }\cos 20^{\circ }-\cos 60^{\circ }\sin 20^{\circ }\right)}{\sin 40^{\circ }}$
$=\frac{4\sin \left(60^{\circ }-20^{\circ }\right)}{\sin 40^{\circ }}$
$[\because \sin (A-B)=\sin A\cos B-\cos A\sin B]$
$=\frac{4\sin 40^{\circ }}{\sin 40^{\circ }}=4$