The value of $\sqrt{3}\cot {20}^0-4\cos {20}^{\circ } \mathrm{is}$

$$\begin{gathered} \Rightarrow \sin {40}^{\circ }=\sin \left({60}^{\circ }-{20}^{\circ }\right) \\ \Rightarrow 2\sin {20}^{\circ }\cos {20}^{\circ }=\frac{\sqrt{3}}{2}\cos {20}^{\circ }-\frac{1}{2}\sin {20}^{\circ } \\ \Rightarrow 4\cos {20}^{\circ }=\frac{\sqrt{3}\cos {20}^{\circ }-\sin {20}^{\circ }}{\sin {20}^{\circ }} \\ \Rightarrow 4\cos {20}^{\circ }=\sqrt{3}\cot {20}^{\circ }-1 \\ \Rightarrow \sqrt{3}\cot {20}^{\circ }-4\cos {20}^{\circ }=1 \end{gathered}$$