Algebra 2 (1st Edition)

$-\dfrac{\sqrt 3}{2}$; $-\dfrac{1}{2}$ and $\sqrt 3$
$\cos a= \dfrac{1}{\sec a}=- \dfrac{1}{\sqrt {1+\tan^2 a}}=-\dfrac{1}{\sqrt {1+(-\sqrt 3)^2}}=-\dfrac{1}{2}$ $\sin 2a=2 \sin a \cos a = 2 (\dfrac{ \sqrt 3}{2}) (-\dfrac{1}{2})=-\dfrac{\sqrt 3}{2}$ $\cos 2a =\cos^2 a-\sin^2 a=(-\dfrac{1}{2})^2 - (\dfrac{ \sqrt 3}{2})^2=\dfrac{1}{4}-\dfrac{3}{4}=-\dfrac{1}{2}$ $\tan 2a =\dfrac{\sin 2a}{\cos 2a}=\dfrac{-\sqrt 3/2}{-1/2}=\sqrt 3$