The value of $\sqrt{3}\text{cosec}20°-\sec 20°$ is

$\sqrt{3}\text{cosec}20°-\sec 20°$

$=(\frac{\sqrt{3}}{\text{sin}20°}-\frac{1}{\cos 20°})$

$=\frac{\sqrt{3}\cos 20°-\sin 20°}{\sin 20°\cos 20°}$

$=\frac{2(\frac{\sqrt{3}}{2}\cos 20°-\frac{1}{2}\sin 20°)}{\frac{1}{2}(2\sin 20°\cos 20°)}$

$=\frac{2(\sin 60°\cos 20°-\cos 60°\sin 20°)}{\frac{1}{2}(\sin 40°)}$

$=4\left[\frac{\sin (60°-20°)}{\sin 40°}\right]$  $\left[\because \sin (A-B)=\sin A\cos B-\cos A\sin B\right]$

$=4\left[\frac{\sin 40°}{\sin 40°}\right]$

$=4$