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, $ y=5; r=13$ $ r=\sqrt {x^2+y^2}$
This gives: $ x=-\sqrt {(13)^2-(5)^2}=-12$; Because $\theta $ lies in Quadrant-II.
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
$\csc \theta =\dfrac{r}{y}=\dfrac{13}{5} \\ \sec \theta =\dfrac{r}{x}=\dfrac{-13}{12} \\ \cot \theta =\dfrac{x}{y}=\dfrac{-12}{5}$
