1. Find the value of $\operatorname{ctg} 10^{\circ}-4 \cos 10^{\circ}=$
\[
\begin{array}{l}
=\frac{\sin 80^{\circ}-2 \sin 20^{\circ}}{\sin 10^{\circ}} \\
=\frac{\sin 80^{\circ}-3 \sin \left(80^{\circ}-60^{\circ}\right)}{\sin 10^{\circ}} \\
=\frac{\sin 80^{\circ}-2\left(\sin 80^{\circ} \cos 60^{\circ}-\cos 80^{\circ} \sin 60^{\circ}\right)}{\sin 10^{\circ}} \\
=\frac{2 \cos 80^{\circ}}{\sin 10^{\circ}} 0^{\circ} \\
=\sqrt{3} \text{. }
\end{array}
\]