4. Find the value: $\frac{\left(\sqrt{3} \cdot \tan 12^{\circ}-3\right) \cdot \csc 12^{\circ}}{4 \cos ^{2} 12^{\circ}-2}$.
4.
$$
\begin{aligned}
\text { Original expression } & =\frac{\left(\frac{\sqrt{3} \sin 12^{\circ}}{\cos 12^{\circ}}-3\right) \cdot \frac{1}{\sin 12^{\circ}}}{4 \cos ^{2} 12^{\circ}-2} \text { (convert tangent and cotangent to sine and cosine) } \\
& =\frac{\sqrt{3} \sin 12^{\circ}-3 \cos 12^{\circ}}{2 \sin 12^{\circ} \cos 12^{\circ}\left(2 \cos ^{2} 12^{\circ}-1\right)} \\
& =\frac{2 \sqrt{3}\left(\frac{1}{2} \sin 12^{\circ}-\frac{\sqrt{3}}{2} \cos 12^{\circ}\right)}{\sin 24^{\circ} \cos 24^{\circ}} \text { (reverse use of double angle formula) } \\
& =\frac{2 \sqrt{3}\left(\sin 12^{\circ} \cos 60^{\circ}-\cos 12^{\circ} \sin 60^{\circ}\right)}{\sin 24^{\circ} \cos 24^{\circ}} \text { (constant transformation) } \\
& =\frac{4 \sqrt{3} \sin \left(12^{\circ}-60^{\circ}\right)}{2 \sin 24^{\circ} \cos 24^{\circ}} \text { (reverse use of difference angle formula) } \\
& =\frac{4 \sqrt{3} \sin \left(-48^{\circ}\right)}{\sin 48^{\circ}}=-4 \sqrt{3} \text { (reverse use of double angle formula). }
\end{aligned}
$$