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

$\begin{array}{c}\cos {290}^{\circ }=\sin {20}^{\circ };\sin {250}^{\circ }=-\sin {70}^{\circ }=-\cos {20}^{\circ }\\ \Rightarrow \frac{1}{\sin {20}^{\circ }}-\frac{1}{\sqrt{3}\cos {20}^{\circ }}\\ =\frac{\sqrt{3}\cos {20}^{\circ }-\sin {20}^{\circ }}{\sqrt{3}\sin {20}^{\circ }\cos {20}^{\circ }}\\ =\frac{2\left[\sin {60}^{\circ }\cos {20}^{\circ }-\sin {20}^{\circ }\cos {60}^{\circ }\right]}{\sqrt{3}\sin {20}^{\circ }\cos {20}^{\circ }}\\ =\frac{4\sin {40}^{\circ }}{\sqrt{3}\sin {40}^{\circ }}=\frac{4\sqrt{3}}{3}\end{array}$