Question

The greatest integer less than or equal to $\frac{1}{\cos {290}^{\circ }}+\frac{1}{\sqrt{3}\sin {250}^{\circ }}$ is

Answer 1

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

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

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

$=\frac{2\sin ({60}^{\circ }-{20}^{\circ })}{\frac{\sqrt{3}}{2}\sin {40}^{\circ }}$

$=\frac{4}{\sqrt{3}}=2.31$

So, greatest integer value of given expression is $2$