The value of  $\sqrt{3}\cot {20}^{\circ }-4\cos {20}^{\circ }$ is

We have,

$\begin{array}{l}\sqrt{3}\cot {20}^{\circ }-4\cos {20}^{\circ }\\ =\frac{\sqrt{3}\cos {20}^{\circ }}{\sin {20}^{\circ }}-4\cos {20}^{\circ }\\ =\frac{\sqrt{3}\cos {20}^{\circ }-4\sin {20}^{\circ }\cos {20}^{\circ }}{\sin {20}^{\circ }}\\ =\frac{2\sin {60}^{\circ }\cos {20}^{\circ }-2\sin {40}^{\circ }}{\sin {20}^{\circ }}\\ =\frac{\sin {80}^{\circ }+\sin {40}^{\circ }-2\sin {40}^{\circ }}{\sin {20}^{\circ }}=\frac{\sin {80}^{\circ }-\sin {40}^{\circ }}{\sin {20}^{\circ }}\end{array}$

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