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

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