Answer
$\cos{\theta} =-\dfrac{5}{13}$
$\tan{\theta}=-\dfrac{12}{5}$
$\csc{\theta} =\dfrac{13}{12}$
$\sec{\theta} =-\dfrac{13}{5}$
$\cot{\theta} =-\dfrac{5}{12}$
Work Step by Step
Since $\theta \in QII$, then $\cos{\theta} <0$.
$\cos{\theta} = -\sqrt{1-\sin^2 {\theta}}$
$\cos{\theta} = -\sqrt{1-\left(\dfrac{12}{13} \right)^2} = -\dfrac{5}{13}$
$\tan{\theta}= \dfrac{\sin{\theta}}{\cos{\theta}}$
$\tan{\theta}= \dfrac{\dfrac{12}{13}}{-\dfrac{5}{13}} = -\dfrac{12}{5}$
$\csc{\theta} = \dfrac{1}{\sin{\theta}}$
$\csc{\theta} = \dfrac{1}{\dfrac{12}{13}}= \dfrac{13}{12}$
$\sec{\theta} = \dfrac{1}{\cos{\theta}}$
$\sec{\theta} = \dfrac{1}{\dfrac{5}{13}} = -\dfrac{13}{5}$
$\cot{\theta} = \dfrac{1}{\tan{\theta}}$
$\cot{\theta} = \dfrac{1}{\dfrac{12}{5}} = -\dfrac{5}{12}$
