Answer
$\sin \theta =\dfrac{y}{r}=\dfrac{2\sqrt {13}}{13} \\ \cos \theta =\dfrac{x}{r}=\dfrac{- 3 \sqrt{13}}{13} \\ \tan \theta =\dfrac{y}{x}=\dfrac{-2}{3}$
and
$\csc \theta =\dfrac{r}{y}=\dfrac{\sqrt {13}}{2} \\ \sec \theta =\dfrac{r}{x}=-\dfrac{\sqrt {13}}{3} \\ \cot \theta =\dfrac{x}{y}=\dfrac{-3}{2}$
Work Step by Step
Here, $ x=-3; y=2$ and $ r=\sqrt {x^2+y^2}$
This gives: $ r=\sqrt {(-3)^2+(2)^2}=\sqrt {13} $; Because $\theta $ lies in Quadrant-II.
The trigonometric ratios are as follows:
$\sin \theta =\dfrac{y}{r}=\dfrac{2\sqrt {13}}{13} \\ \cos \theta =\dfrac{x}{r}=\dfrac{- 3 \sqrt{13}}{13} \\ \tan \theta =\dfrac{y}{x}=\dfrac{-2}{3}$
and
$\csc \theta =\dfrac{r}{y}=\dfrac{\sqrt {13}}{2} \\ \sec \theta =\dfrac{r}{x}=-\dfrac{\sqrt {13}}{3} \\ \cot \theta =\dfrac{x}{y}=\dfrac{-3}{2}$
