Question

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

• A. $\frac{2\sqrt{3}}{3}$
• B. $\frac{4\sqrt{3}}{3}$
• C. $\sqrt{3}$
• D. none

Answer 1

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

Given,

$\frac{1}{\cos 290}+\frac{1}{\sqrt{3}\sin 250}$

$=\frac{1}{\cos (270+20)}+\frac{1}{\sqrt{3}\sin (270-20)}$

$=\frac{1}{\sin 20}-\frac{1}{\sqrt{3}\cos 20}$

$=\frac{\sqrt{3}\cos 20-\sin 20}{\sin 20\sqrt{3}\cos 20}$

$=\frac{2(\frac{\sqrt{3}}{2}\cos 20-\frac{1}{2}\sin 20)}{\frac{1}{2}\sqrt{3}\sin 40}$

$=\frac{4}{\sqrt{3}}\frac{(\sin 60\cos 20-\cos 60\sin 20)}{\sin 40}$

$=\frac{4}{\sqrt{3}}\frac{\sin (60-20)}{\sin 40}$

$=\frac{4}{\sqrt{3}}\frac{\sin 40}{\sin 40}$

$=\frac{4}{\sqrt{3}}$

$=\frac{4\sqrt{3}}{3}$