2015/098) Prove $(4 \cos\, 20^\circ + 1) \tan 20^\circ = \sqrt3$

we have $tan\, 60^\circ – \tan\, 20^\circ$
= $\dfrac{\sin\, 60^\circ}{\cos\,60^\circ} – \dfrac{\sin\, 20^\circ}{\cos\,20^\circ}$
= $\dfrac{\sin\, 60 ^\circ\cos\, 20^\circ – \cos\, 60^\circ \sin\, 20^\circ}{\cos\, 60^\circ\, \cos\, 20^\circ}$
= $\dfrac{\sin\, 40^\circ}{\cos\, 60^\circ \cos\, 20^\circ}$
=  $2\dfrac{\sin\, 40^\circ}{\cos\, 20^\circ}$
= $2 \dfrac{2 sin\, 20^\circ \cos\, 20^\circ}{\cos\, 20^\circ} = 4 \sin\, 20^\circ$
hence $4\sin \, 20^\circ + \tan\, 20^\circ = \sqrt3$
or $4 \cos\, 20^\circ \tan\, 20^\circ + \tan \,20^\circ = \sqrt3$
or $(4 \cos\, 20^\circ + 1) \tan\, 20^\circ = \sqrt3$