Answer
$\dfrac{4}{5}$ and $\dfrac{-3}{5}$ and $-\dfrac{4}{3}$
Work Step by Step
$\cos a= \dfrac{1}{\sec a}=- \dfrac{1}{\sqrt {1+\tan^2 a}}=-\dfrac{1}{\sqrt {1+2^2}}=-\dfrac{\sqrt 5}{5}$
$\sin 2a=2 \sin a \cos a = 2 (-\dfrac{ 2\sqrt 5}{5}) (-\dfrac{\sqrt 5}{5})=\dfrac{4}{5}$
$\cos 2a =\cos^2 a-\sin^2 a=(-\dfrac{\sqrt 5}{5})^2 - (-\dfrac{ 2\sqrt 5}{5})^2=\dfrac{1}{5}-\dfrac{4}{5}=\dfrac{-3}{5}$
$\tan 2a =\dfrac{\sin 2a}{\cos 2a}=\dfrac{4/5}{-3/5}=-\dfrac{4}{3}$
