$\sqrt{3}\mathrm{cosec}{20}^{\circ }-\sec {20}^{\circ }$ is equal to

$\begin{array}{l}\sqrt{3}\mathrm{cosec}{20}^{\circ }-\sec {20}^{\circ }\\ =\frac{\sqrt{3}}{\sin {20}^{\circ }}-\frac{1}{\cos {20}^{\circ }}=\frac{\sqrt{3}\cos {20}^{\circ }-\sin {20}^{\circ }}{\sin {20}^{\circ }\cdot \cos {20}^{\circ }}\\ =\frac{4\left(\frac{\sqrt{3}}{2}\cos {20}^{\circ }-\frac{1}{2}\sin {20}^{\circ }\right)}{2\sin {20}^{\circ }\cdot \cos {20}^{\circ }}\\ =\frac{4\left(\sin {60}^{\circ }\cos {20}^{\circ }-\cos {60}^{\circ }\sin {20}^{\circ }\right)}{\sin {40}^{\circ }}\\ =\frac{4\sin {40}^{\circ }}{\sin {40}^{\circ }}=4\end{array}$