what is the value

$\begin{aligned} \sqrt 3\cot 20^\circ - 4 \cos 20^\circ & = \frac \blue{2\cos 30^\circ \cos 20^\circ}{\sin 20^\circ} - 4\cos 20^\circ & \small \blue{\text{Note that }2\cos A \cos B = \cos (A-B) + \cos (A+B)} \\ & = \frac \blue{\cos 10^\circ + \cos 50^\circ}{\sin 20^\circ} - 4\cos 20^\circ & \small \blue{\text{and }\cos \theta = \sin (90^\circ - \theta)} \\ & = \frac \blue{\sin 80^\circ + \sin 40^\circ}{\sin 20^\circ} - 4\cos 20^\circ & \small \blue{\text{and }\sin 2\theta = 2 \sin \theta \cos \theta} \\ & = \frac {\sin 40^\circ(2\cos 40^\circ + 1)}{\sin 20^\circ} - 4\cos 20^\circ \\ & = 2\cos 20^\circ(2 \cos 40^\circ + 1) - 4\cos 20^\circ \\ & = 4 \cos 20^\circ \cos 40^\circ - 2\cos 20^\circ \\ & = 2 \cos 20^\circ + 2 \cos 60^\circ - 2\cos 20^\circ \\ & = \boxed 1 \end{aligned}$

$\sin 40\degree=\sin (60\degree-20\degree)\implies 2\sin 20\degree\cos 20\degree=$

$\sin 60\degree\cos 20\degree-\cos 60\degree\sin 20\degree$

$\implies 4\cos 60\degree\sin 20\degree\cos 20\degree=\sin 60\degree\cos 20\degree-\cos 60\degree\sin 20\degree$

$\implies \dfrac{\sin 60\degree\cos 20\degree-\cos 60\degree\sin 20\degree}{\cos 20\degree\cos 60\degree}=4\sin 20\degree$

$\implies \tan 60\degree-\tan 20\degree=4\sin 20\degree$

$\implies \tan 20\degree=\sqrt 3-4\sin 20\degree$

$\implies \sqrt 3\cot 20\degree-4\cos 20\degree=$

$\cot 20\degree(\sqrt 3-4\sin 20\degree)=\cot 20\degree\times \tan 20\degree=\boxed 1$ .