Question

The value of the expression $\frac{1}{\cos {290}^o}+\frac{1}{\sqrt{3}\sin {250}^o}$ is equal to

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

Answer 1

Answer: (B)

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

$\frac{1}{\cos {290}^0}+\frac{1}{\sqrt{3}\sin {250}^0}$
$=\frac{1}{\cos ({270}^o+{20}^o)}+\frac{1}{\sqrt{3}\sin ({270}^o-{20}^o)}$
$=\frac{1}{\sin {20}^o}+\frac{1}{\sqrt{3}(-\cos {20}^o)}$
$=\frac{\sqrt{3}\cos {20}^o-\sin {20}^o}{\sqrt{3}\sin {20}^o\cos {20}^o}$
$=\frac{\frac{\sqrt{3}}{2}\cos {20}^o-\frac{1}{2}\sin {20}^o}{\frac{\sqrt{3}}{4}\sin {40}^o}$
$=\frac{\sin {60}^o\cos {20}^o-\cos {60}^o\sin {20}^o}{\frac{\sqrt{3}}{4}\sin {40}^o}$
$=\frac{\sin {40}^o}{\frac{\sqrt{3}}{4}\sin {40}^o}=\frac{4}{\sqrt{3}}$.