Answer: [tex]\bold{\dfrac{3}{4}}[/tex]
Step-by-step explanation:
Use the Unit Circle to find the values of each trig term (see attachment)
[tex]\sin\ 300^o=-\dfrac{\sqrt3}{2}\\\\\\\cos\ 390^o\ \text{is one rotation of}\ 360^o+30^o\rightarrow\cos\ 30^o=\dfrac{\sqrt3}{2}\\\\\\\cot\ 150^o=\dfrac{\cos\ 150^o}{\sin\ 150^o}=\dfrac{-\sqrt3/2}{1/2}=-\sqrt3\\\\\\\sin\ 120^o=\dfrac{\sqrt3}{2}[/tex]
[tex].\quad (\sin\ 300^o)(\cos\ 390^o)-(\cot\ 150^o)(\sin\ 120^o)\\\\=\quad \bigg(\dfrac{-\sqrt3}{2}\bigg)\bigg(\dfrac{\sqrt3}{2}\bigg)-\bigg(\dfrac{-\sqrt3}{1}\bigg)\bigg(\dfrac{\sqrt3}{2}\bigg)\\\\\\=\qquad \quad \qquad -\dfrac{3}{4}\quad -\quad \bigg(-\dfrac{3}{2}\bigg)\\\\\\=\qquad \qquad \quad -\dfrac{3}{4}\quad +\qquad \dfrac{6}{4}\\\\\\=\qquad \qquad \qquad \qquad \large\boxed{\dfrac{3}{4}}[/tex]
