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