[tex]\sec x=\dfrac{1}{\cos x}\\\\\sec\theta=\dfrac{2\sqrt3}{3} \ \textgreater \ 0\ and\ \sin\theta \ \textless \ 0\to\theta\ is\ in\ the\ 4-th\ quadrant.\\\theta\in\left(\dfrac{3\pi}{2};\ 2\pi\right)\\\\\sec\theta=\dfrac{2\sqrt3}{3}\\\\\dfrac{1}{\cos\theta}=\dfrac{2\sqrt3}{3}\Rightarrow\cos\theta=\dfrac{3}{2\sqrt3}\cdot\dfrac{\sqrt3}{\sqrt3}\\\\\cos\theta=\dfrac{3\sqrt3}{2\cdot3}\\\\\cos\theta=\dfrac{\sqrt3}{2}\to\theta=\dfrac{11\pi}{6}\\\\\cot\theta=\cot\frac{11\pi}{6}=\cot\frac{5\pi}{6}=-\sqrt3[/tex]
[tex]Answer:\boxed{\sqrt2-\sqrt3}[/tex]
