We have that \begin{align*} \tan 100^\circ + 4 \sin 100^\circ &= \frac{\sin 100^\circ}{\cos 100^\circ} + 4 \sin 100^\circ \\ &= \frac{\sin 80^\circ}{-\cos 80^\circ} + 4 \sin 80^\circ \\ &= -\frac{\cos 10^\circ}{\sin 10^\circ} + 4 \cos 10^\circ \\ &= \frac{4 \cos 10^\circ \sin 10^\circ - \cos 10^\circ}{\sin 10^\circ}.\end{align*}By double angle formula, \begin{align*} \frac{4 \cos 10^\circ \sin 10^\circ - \cos 10^\circ}{\sin 10^\circ} &= \frac{2 \sin 20^\circ - \cos 10^\circ}{\sin 10^\circ} \\ &= \frac{\sin 20^\circ + \sin 20^\circ - \sin 80^\circ}{\sin 10^\circ}.\end{align*}By sum-to-product, \[\sin 20^\circ - \sin 80^\circ = 2 \cos 50^\circ \sin (-30^\circ) = -\cos 50^\circ,\]so \begin{align*} \frac{\sin 20^\circ + \sin 20^\circ - \sin 80^\circ}{\sin 10^\circ} &= \frac{\sin 20^\circ - \cos 50^\circ}{\sin 10^\circ} \\ &= \frac{\sin 20^\circ - \sin 40^\circ}{\sin 10^\circ}.\end{align*}By sum-to-product, \[\sin 20^\circ - \sin 40^\circ = 2 \cos 30^\circ \sin (-10^\circ) = -\sqrt{3} \sin 10^\circ,\]so $ \frac{\sin 20^\circ - \sin 40^\circ}{\sin 10^\circ} = \boxed{-\sqrt{3}} $.