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