Question

Find the value of sin 300°.cos 390°-cot 150°.sin 120°​

Answer 1

Answer:  $\bold{\dfrac{3}{4}}$

Step-by-step explanation:

Use the Unit Circle to find the values of each trig term (see attachment)

$\sin\ 300^o=-\dfrac{\sqrt3}{2}\\\\\\\cos\ 390^o\ \text{is one rotation of}\ 360^o+30^o\rightarrow\cos\ 30^o=\dfrac{\sqrt3}{2}\\\\\\\cot\ 150^o=\dfrac{\cos\ 150^o}{\sin\ 150^o}=\dfrac{-\sqrt3/2}{1/2}=-\sqrt3\\\\\\\sin\ 120^o=\dfrac{\sqrt3}{2}$

$.\quad (\sin\ 300^o)(\cos\ 390^o)-(\cot\ 150^o)(\sin\ 120^o)\\\\=\quad \bigg(\dfrac{-\sqrt3}{2}\bigg)\bigg(\dfrac{\sqrt3}{2}\bigg)-\bigg(\dfrac{-\sqrt3}{1}\bigg)\bigg(\dfrac{\sqrt3}{2}\bigg)\\\\\\=\qquad \quad \qquad -\dfrac{3}{4}\quad -\quad \bigg(-\dfrac{3}{2}\bigg)\\\\\\=\qquad \qquad \quad -\dfrac{3}{4}\quad +\qquad \dfrac{6}{4}\\\\\\=\qquad \qquad \qquad \qquad \large\boxed{\dfrac{3}{4}}$

Answer 2

note that :

cot150 = 1/tan150

cos390 = cos(360+30) = cos30