Answer
$\tan{\theta} =\sqrt{3}$
$\cos{\theta} = -\dfrac{1}{2}$
$\sin{\theta}=-\dfrac{\sqrt{3}}{2}$
$\csc{\theta} =-\dfrac{2\sqrt{3}}{3}$
$\cot{\theta} = \dfrac{\sqrt{3}}{3}$
Work Step by Step
$1+\tan^2 {\theta} = \sec^2{\theta}$
$\tan^2 {\theta} = \sec^2{\theta}-1$
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,
$\csc{\theta} = \dfrac{1}{-\dfrac{\sqrt{3}}{2}}= -\dfrac{2\sqrt{3}}{3}$
$\cot{\theta} = \dfrac{1}{\tan{\theta}}$
$\cot{\theta} = \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}$
