Answer
$\cos\theta=-\dfrac{5}{13}$
$\tan\theta=-\dfrac{12}{5}$
$\cot\theta=-\dfrac{5}{12}$
$\sec\theta=-\dfrac{13}{5}$
$\csc\theta=\dfrac{13}{12}$
Work Step by Step
$\sin\theta=\dfrac{12}{13}=\dfrac{y}{r}$
Thus,
$y=12$
$r=13$
We find $x$ using the formula $r^2=x^2+y^2$:
$r^2=x^2+y^2$
$13^2=x^2+12^2$
$169=x^2+144$
$x^2=25$
$x=\pm\sqrt{25}$
$x=\pm5$
Since the angle is in $\text{Quadrant II}$ where $x$ is negative, then $x=-5$.
Compute for $\cos\theta, \tan\theta, \cot\theta,\sec\theta, \csc\theta$:
$\cos\theta=\dfrac{x}{r}=\dfrac{-5}{13}=-\dfrac{5}{13}$
$\tan\theta=\dfrac{y}{x}=\dfrac{12}{-5}=-\dfrac{12}{5}$
$\cot\theta=\dfrac{x}{y}=\dfrac{-5}{12}=-\dfrac{5}{12}$
$\sec\theta=\dfrac{r}{x}=\dfrac{13}{-5}=-\dfrac{13}{5}$
$\csc\theta=\dfrac{r}{y}=\dfrac{13}{12}$
