Question

The value of $\frac{1}{\cos {290}^{\circ }}+\frac{1}{\sqrt{3}\sin {250}^{\circ }}$ is

• A. $\frac{\sqrt{3}}{4}$
• B. $\frac{\sqrt{5}}{2}$
• C. $\frac{4}{\sqrt{3}}$
• D. $\frac{4}{\sqrt{5}}$

Answer 1

The correct option is C $\frac{4}{\sqrt{3}}$

$\cos {290}^{\circ }=\cos ({270}^{\circ }+{20}^{\circ })=\sin {20}^{\circ }$
$\sin {250}^{\circ }=\sin ({270}^{\circ }-{20}^{\circ })=-\cos {20}^{\circ }$

$\frac{1}{\cos {290}^{\circ }}+\frac{1}{\sqrt{3}\sin {250}^{\circ }}$
$=\frac{1}{\sin {20}^{\circ }}-\frac{1}{\sqrt{3}\cos {20}^{\circ }}$
$=\frac{\sqrt{3}\cos {20}^{\circ }-\sin {20}^{\circ }}{\sqrt{3}\cos {20}^{\circ }\sin {20}^{\circ }}=\frac{2[\frac{\sqrt{3}}{2}\cos {20}^{\circ }-\frac{1}{2}\sin {20}^{\circ }]}{\sqrt{3}\cos {20}^{\circ }\sin {20}^{\circ }}$
$=\frac{2[\sin {60}^{\circ }\cos {20}^{\circ }-\cos {60}^{\circ }\sin {20}^{\circ }]}{\sqrt{3}\cos {20}^{\circ }\sin {20}^{\circ }}$
$=\frac{4\sin ({60}^{\circ }-{20}^{\circ })}{2\sqrt{3}\cos {20}^{\circ }\sin {20}^{\circ }}=\frac{4\sin {40}^{\circ }}{\sqrt{3}\sin {40}^{\circ }}(\because 2\sin \theta \cos \theta =\sin 2\theta )=\frac{4}{\sqrt{3}}$