Answer
$\sin{\theta} =-\dfrac{3\sqrt{13}}{13}$
$\cos{\theta} =\dfrac{2\sqrt{13}}{13}$
$\tan{\theta} =-\dfrac{3}{2}$
$\csc{\theta} =-\dfrac{\sqrt{13}}{13}$
$\sec{\theta} =\dfrac{\sqrt{13}}{2}$
$\cot{\theta} =-\dfrac{2}{3} $
Work Step by Step
$\cos{\theta} = \dfrac{2\sqrt{13}}{13}$
$\because \theta \in QIV \hspace{20pt} \therefore \sin{\theta}$ is negative.
$\sin{\theta} = - \sqrt{1-\cos^2{\theta}} = - \sqrt{1-\left(\dfrac{2\sqrt{13}}{13}\right)^2} = -\dfrac{3\sqrt{13}}{13}$
$\tan{\theta} = \dfrac{\sin{\theta}}{\cos{\theta}} = -\dfrac{3}{2}$
$\csc{\theta} = \dfrac{1}{\sin{\theta}} = -\dfrac{\sqrt{13}}{13}$
$\sec{\theta} = \dfrac{1}{\cos{\theta}} = \dfrac{\sqrt{13}}{2}$
$\cot{\theta} = \dfrac{1}{\tan{\theta}} = -\dfrac{2}{3} $
