Answer
$\sin \theta =\dfrac{y}{r}=\dfrac{-2 \sqrt 2}{3} \\ \cos \theta =\dfrac{x}{r}=\dfrac{-1}{3} \\ \tan \theta =\dfrac{y}{x}=2 \sqrt 2$
and
$\csc \theta =\dfrac{r}{y}=\dfrac{-3 \sqrt 2}{4} \\ \sec \theta =\dfrac{r}{x}=-\dfrac{3}{2\sqrt 2}=-\dfrac{3\sqrt 2}{4} \\ \cot \theta =\dfrac{x}{y}=\dfrac{\sqrt 2}{4}$
Work Step by Step
Here, $ x=-1; r=3$ and $ r=\sqrt {x^2+y^2}$
This gives: $ y=-\sqrt {(3)^2-(-1)^2}=-2 \sqrt 2$; Because $\theta $ lies in Quadrant-III.
The trigonometric ratios are as follows:
$\sin \theta =\dfrac{y}{r}=\dfrac{-2 \sqrt 2}{3} \\ \cos \theta =\dfrac{x}{r}=\dfrac{-1}{3} \\ \tan \theta =\dfrac{y}{x}=2 \sqrt 2$
and
$\csc \theta =\dfrac{r}{y}=\dfrac{-3 \sqrt 2}{4} \\ \sec \theta =\dfrac{r}{x}=-\dfrac{3}{2\sqrt 2}=-\dfrac{3\sqrt 2}{4} \\ \cot \theta =\dfrac{x}{y}=\dfrac{\sqrt 2}{4}$
