Question

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

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

Answer 1

Answer: (B)

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

$\frac{1}{\cos {290}^o}+\frac{1}{\sqrt{3}\sin {250}^o}=\frac{1}{\cos ({270}^o+{20}^o)}+\frac{1}{\sqrt{3}\sin ({270}^o-{20}^o)}$
$=\frac{1}{\sin {20}^o}+\frac{1}{-\sqrt{3}\cos {20}^o}=\frac{\sqrt{3}\cos {20}^o-\sin {20}^o}{\sqrt{3}\cos {20}^o\sin {20}^o}$
$=\frac{2(\frac{\sqrt{3}}{2}\cos {20}^o-\frac{1}{2}\sin {20}^o)}{\frac{\sqrt{3}}{2}\sin {40}^o}=\frac{2\sin ({60}^o-{20}^o)}{\frac{\sqrt{3}}{2}\sin {40}^o}=\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3}$