Trigonometry (10th Edition)

$\sqrt3$
RECALL: $\tan{x} = \dfrac{\sin{x}}{\cos{x}}$ The angle $\frac{\pi}{3}$ is a special angle and the following values are known: $\sin{\frac{\pi}{3}}=\frac{\sqrt3}{2}$ $\cos{\frac{\pi}{3}}=\frac{1}{2}$ Thus, $\tan{\frac{\pi}{3}} = \dfrac{\sin{\frac{\pi}{3}}}{\cos{\frac{\pi}{3}}}=\dfrac{\frac{\sqrt3}{2}}{\frac{1}{2}}=\frac{\sqrt3}{2} \cdot \frac{2}{1}=\sqrt3$