Answer
$\sin \theta =\dfrac{y}{r}=\dfrac{-4}{5} \\ \cos \theta =\dfrac{x}{r}=\dfrac{-3}{5} \\ \tan \theta =\dfrac{y}{x}=\dfrac{4}{3}$
and
$\csc \theta =\dfrac{r}{y}=\dfrac{-5}{4} \\ \sec \theta =\dfrac{r}{x}=\dfrac{-5}{3} \\ \cot \theta =\dfrac{x}{y}=\dfrac{3}{4}$
Work Step by Step
Here, $ x= -3; r=5$ $ r=\sqrt {x^2+y^2}$
This gives: $ y=-\sqrt {(5)^2-(-3)^2}=-4$; Because $\theta $ lies in Quadrant-III.
The trigonometric ratios are as follows:
$\sin \theta =\dfrac{y}{r}=\dfrac{-4}{5} \\ \cos \theta =\dfrac{x}{r}=\dfrac{-3}{5} \\ \tan \theta =\dfrac{y}{x}=\dfrac{4}{3}$
and
$\csc \theta =\dfrac{r}{y}=\dfrac{-5}{4} \\ \sec \theta =\dfrac{r}{x}=\dfrac{-5}{3} \\ \cot \theta =\dfrac{x}{y}=\dfrac{3}{4}$
