## Precalculus (6th Edition) Blitzer

$\dfrac{1}{2 \sqrt 3}$ or, $\dfrac{\sqrt 3}{6}$
Solve. $\tan ( \dfrac{\pi}{6}) \times \cos ( \dfrac{\pi}{3})- \cos \dfrac{\pi}{2}$ $=\dfrac{1}{\sqrt 3} \times \dfrac{1}{2} -0$ $= \dfrac{1}{2 \sqrt 3} \times \dfrac{\sqrt 3}{ \sqrt 3}$ $= \dfrac{\sqrt 3}{6}$