Consider $\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}=\frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}}$
$=\frac{\sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$
Multiply and divide by 2 in $N^{r}$.
$=2\left(\frac{\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}\right)$
$=2\left(\frac{\sin 60^{\circ} \cos 20^{\circ}-\cos 60^{\circ} \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}\right)$
$\left(\because \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right.$ and $\left.\cos 60^{\circ}=\frac{1}{2}\right)$
$=\frac{2 \times 2\left[\sin \left(60^{\circ}-20^{\circ}\right)\right]}{2 \sin 20^{\circ} \cos 20^{\circ}}$
$(\because \sin A \cos B-\cos A \sin B=\sin (A-B)$ and $\sin 2 \theta=$
$2 \sin \theta \cos \theta)$
$=\frac{4 \sin 40^{\circ}}{\sin 40^{\circ}}=4$