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

We have,

$\begin{array}{l}\sqrt{3}\stackrel{ }{ }\mathrm{cosec}{20}^{\circ }-\sec {20}^{\circ }\\ =\tan {60}^{\circ }\mathrm{cosec}{20}^{\circ }-\sec {20}^{\circ }\\ =\frac{\sin {60}^{\circ }\cos {20}^{\circ }-\cos {60}^{\circ }\sin {20}^{\circ }}{\cos {60}^{\circ }\sin {20}^{\circ }\cos {20}^{\circ }}\end{array}$

$=\frac{\sin {40}^{\circ }}{\cos {60}^{\circ }\sin {20}^{\circ }\cos {20}^{\circ }}=\frac{2\sin {20}^{\circ }\cos {20}^{\circ }}{1}=4$