## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\dfrac{4}{3}$
We have: $\sin \theta=\dfrac{2}{\sqrt 5}$ and $\cos \theta=-\dfrac{1}{\sqrt 5}$ Since, $\tan (\theta)=\dfrac{\sin \theta}{\cos \theta} \implies \tan \theta=\dfrac{2/\sqrt 5}{-1/\sqrt 5}=-2$ We plug these values into the following identity: $\tan (2 \ \theta)=\dfrac{2 \tan \theta}{ 1-\tan^2 \theta} = \dfrac{(2)(-2) }{ 1- (-2)^2}=\dfrac{4}{3}$