Question

The value of the expression $\frac{1}{\sqrt{3}\sin \left(250^{\circ }\right)}+\frac{1}{\cos \left(290^{\circ }\right)}$ is

• A. $\frac{\sqrt{3}}{4}$
• B. $\frac{4}{\sqrt{3}}$
• C. $\sqrt{3}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

Answer: (B)

$E=\frac{1}{\sin 20^{\circ }}-\frac{1}{\sqrt{3}\cos 20^{\circ }}=\frac{\frac{\sqrt{3}}{2}\cos 20^{\circ }-\frac{1}{2}\sin 20^{\circ }}{\frac{\sqrt{3}}{2}\sin 20^{\circ }\cos 20^{\circ }}=\frac{\sin \left(60^{\circ }-20^{\circ }\right)}{\frac{\sqrt{3}}{4}\cdot \sin 40^{\circ }}$