It is given that
[tex]\sec (\theta)=\frac{2\sqrt[]{3}}{3}[/tex]It can be written as follows:
[tex]\sec (\theta)=\frac{2\sqrt[]{3}}{\sqrt[]{3}\sqrt[]{3}}[/tex][tex]\sec (\theta)=\frac{2}{\sqrt[]{3}}[/tex][tex]\text{Substitute sec}\theta=\frac{1}{\cos \theta}[/tex][tex]\frac{1}{\cos \theta}=\frac{2}{\sqrt[]{3}}[/tex]Reciprocal, we get
[tex]\cos \theta=\frac{\sqrt[]{3}}{2}[/tex]We know that
[tex]\cos 30^o=\frac{\sqrt[]{3}}{2}[/tex][tex]\cos (2\pi-\theta)=\cos \theta[/tex][tex]\text{Substitute }\theta=30^o,\text{ we get}[/tex][tex]\cos (360^o-30^o)=\cos 30^o[/tex][tex]\cos 330^o=\cos 30^o=\frac{\sqrt[]{3}}{2}[/tex]Hence we get
[tex]\sec 330^o=\sec 30^o=\frac{2\sqrt[]{3}}{3}[/tex]