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)

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