Answer
$\sin \theta =\dfrac{y}{r}=\dfrac{-5}{13} \\ \cos \theta =\dfrac{x}{r}=\dfrac{-12}{13} \\ \tan \theta =\dfrac{y}{x}=\dfrac{5}{12}$
and
$\csc \theta =\dfrac{r}{y}=\dfrac{-13}{5} \\ \sec \theta =\dfrac{r}{x}=-\dfrac{13}{12} \\ \cot \theta =\dfrac{x}{y}=\dfrac{12}{5}$
Work Step by Step
Here, $ x=-12; y=-5$ and $ r=\sqrt {x^2+y^2}$
This gives: $ r=\sqrt {(-12)^2+(-5)^2}=13$; Because $\theta $ lies in Quadrant-III.
The trigonometric ratios are as follows:
$\sin \theta =\dfrac{y}{r}=\dfrac{-5}{13} \\ \cos \theta =\dfrac{x}{r}=\dfrac{-12}{13} \\ \tan \theta =\dfrac{y}{x}=\dfrac{5}{12}$
and
$\csc \theta =\dfrac{r}{y}=\dfrac{-13}{5} \\ \sec \theta =\dfrac{r}{x}=-\dfrac{13}{12} \\ \cot \theta =\dfrac{x}{y}=\dfrac{12}{5}$
