Answer
$ \sin{\theta} = \dfrac{5}{13}$
$\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
$ \csc{\theta} = \dfrac{r}{y} = \dfrac{13}{5}$
$\therefore y = 5 \hspace{20pt} r = 13$
$\cos{\theta} = \dfrac{x}{r} \hspace{20pt} \because \cos{\theta} <0 \hspace{20pt}\therefore x $ is negative.
$x = -\sqrt{r^2-y^2} = -\sqrt{(13)^2-(5)^2} = -12$
$\sin{\theta} = \dfrac{y}{r} = \dfrac{5}{13}$
$\cos{\theta} = \dfrac{x}{r} = -\dfrac{12}{13} $
$\tan{\theta} = \dfrac{y}{x} = -\dfrac{5}{12}$
$\sec{\theta} = \dfrac{1}{\cos{\theta}} = -\dfrac{13}{12}$
$\cot{\theta} = \dfrac{1}{\tan{\theta}} = -\dfrac{12}{5}$
