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
$\csc \theta =\dfrac{r}{y}=\dfrac{3}{-2\sqrt 2}=\dfrac{-3\sqrt 2}{4} \\ \sec \theta =\dfrac{r}{x}=3 \\ \cot \theta =\dfrac{x}{y}=\dfrac{-\sqrt 2}{4}$
Work Step by Step
Here, $ x=1; r=3$ $ r=\sqrt {x^2+y^2}$
This gives: $ y=-\sqrt {(3)^2-(1)^2}=-2\sqrt 2$; Because $\theta $ lies in Quadrant-IV.
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
$\csc \theta =\dfrac{r}{y}=\dfrac{3}{-2\sqrt 2}=\dfrac{-3\sqrt 2}{4} \\ \sec \theta =\dfrac{r}{x}=3 \\ \cot \theta =\dfrac{x}{y}=\dfrac{-\sqrt 2}{4}$
