Answer
$\cos{\theta} =-\dfrac{12}{13}$
$\tan{\theta}=-\dfrac{5}{12}$
$\csc{\theta} =\dfrac{13}{5}$
$\sec{\theta} = -\dfrac{13}{12}$
$\cot{\theta} =-\dfrac{12}{5}$
Work Step by Step
Since $90^{\circ} < \theta <180^{\circ}$, then $\theta \in QII$ and $\cos{\theta} <0$.
Thus,
$\cos{\theta} = -\sqrt{1-\sin^2 {\theta}}$
$\cos{\theta} = -\sqrt{1-\left(\dfrac{5}{13} \right)^2} = -\dfrac{12}{13}$
$\tan{\theta}= \dfrac{\sin{\theta}}{\cos{\theta}}$
$\tan{\theta}= \dfrac{\dfrac{5}{13}}{-\dfrac{12}{13}} = -\dfrac{5}{12}$
$\csc{\theta} = \dfrac{1}{\sin{\theta}}$
$\csc{\theta} = \dfrac{1}{\dfrac{5}{13}}= \dfrac{13}{5}$
$\sec{\theta} = \dfrac{1}{\cos{\theta}}$
$\sec{\theta} = \dfrac{1}{-\dfrac{12}{13}} = -\dfrac{13}{12}$
$\cot{\theta} = \dfrac{1}{\tan{\theta}}$
$\cot{\theta} = \dfrac{1}{-\dfrac{5}{12}} = -\dfrac{12}{5}$
