Solving $E=\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}$

Divide both terms by two and use the fact $\sin(30) = \frac{1}{2}$ and $\cos(30) = \frac{\sqrt{3}}{2}$. Then you just need to use the formulas for $\sin(a+b)$ and $\sin(a-b)$ to find the solution.

We have \begin{eqnarray*} E&=&\frac{\cos 10^\circ-\sqrt{3}\sin 10^\circ}{\sin 10^\circ \cos 10^\circ}\\ &=&4\frac{(1/2)\cos 10^\circ-(\sqrt{3}/2)\sin 10^\circ}{2\sin 10^\circ \cos 10^\circ}\\ &=&4\frac{\cos 60^\circ\cos 10^\circ-\sin 60^\circ\sin 10^\circ}{\sin 20^\circ}\\ &=&4\frac{\cos(60^\circ+10^\circ)}{\sin 20^\circ}\\ &=&4\frac{\cos 70^\circ}{\sin 20^\circ}\\ &=&4\frac{\sin 20^\circ}{\sin 20^\circ}\\ &=&4. \end{eqnarray*}