We are given the expression $\csc 10^{\circ} - \sqrt{3} \sec 10^{\circ}$.

We can rewrite this as:

$\frac{1}{\sin 10^{\circ}} - \frac{\sqrt{3}}{\cos 10^{\circ}}$

Combining the fractions, we get:

$\frac{\cos 10^{\circ} - \sqrt{3} \sin 10^{\circ}}{\sin 10^{\circ} \cos 10^{\circ}}$

Multiplying and dividing by 2, we have:

$\frac{2(\frac{1}{2} \cos 10^{\circ} - \frac{\sqrt{3}}{2} \sin 10^{\circ})}{\sin 10^{\circ} \cos 10^{\circ}}$

We know that $\sin 30^{\circ} = \frac{1}{2}$ and $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$. So, we can rewrite the expression as:

$\frac{2(\sin 30^{\circ} \cos 10^{\circ} - \cos 30^{\circ} \sin 10^{\circ})}{\sin 10^{\circ} \cos 10^{\circ}}$

Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$, we get:

$\frac{2 \sin(30^{\circ} - 10^{\circ})}{\sin 10^{\circ} \cos 10^{\circ}} = \frac{2 \sin 20^{\circ}}{\sin 10^{\circ} \cos 10^{\circ}}$

Multiplying and dividing the denominator by 2, we have:

$\frac{4 \sin 20^{\circ}}{2 \sin 10^{\circ} \cos 10^{\circ}}$

Using the identity $\sin 2A = 2 \sin A \cos A$, we get:

$\frac{4 \sin 20^{\circ}}{\sin 20^{\circ}}$

Therefore, the expression simplifies to:

$4$