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

$\tan{\theta} =\sqrt{3}$ $\cos{\theta} = -\dfrac{1}{2}$ $\sin{\theta}=-\dfrac{\sqrt{3}}{2}$ $\csc{\theta} =-\dfrac{2\sqrt{3}}{3}$ $\cot{\theta} = \dfrac{\sqrt{3}}{3}$
$1+\tan^2 {\theta} = \sec^2{\theta}$ $\tan^2 {\theta} = \sec^2{\theta}-1$ Since $\tan{\theta} >0$, then: $\tan{\theta} = \sqrt{\sec^2{\theta}-1}$ Thus, $\tan{\theta} = \sqrt{(-2)^2-1} = \sqrt{3}$ With $\cos{\theta} = \dfrac{1}{\sec{\theta}}$, then $\cos{\theta} =\dfrac{1}{-2} = -\dfrac{1}{2}$ Since $\sin{\theta} = \cos{\theta} \tan{\theta}$, then $\sin{\theta}= -\dfrac{1}{2} \times \sqrt{3} = -\dfrac{\sqrt{3}}{2}$ Recall that $\csc{\theta} = \dfrac{1}{\sin{\theta}}$. Hence, $\csc{\theta} = \dfrac{1}{-\dfrac{\sqrt{3}}{2}}= -\dfrac{2\sqrt{3}}{3}$ $\cot{\theta} = \dfrac{1}{\tan{\theta}}$ $\cot{\theta} = \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}$