Answer
$\sin \theta =\dfrac{y}{r}=\dfrac{-15}{17} \\ \cos \theta =\dfrac{x}{r}=\dfrac{8}{17} \\ \tan \theta =\dfrac{y}{x}=\dfrac{-15}{8}$
and
$\csc \theta =\dfrac{r}{y}=\dfrac{-17}{15} \\ \sec \theta =\dfrac{r}{x}=\dfrac{17}{8} \\ \cot \theta =\dfrac{x}{y}=\dfrac{-8}{15}$
Work Step by Step
Here, $ x=8; r=17$ $ r=\sqrt {x^2+y^2}$
This gives: $ y=-\sqrt {(17)^2-(8)^2}=-15$; Because $\theta $ lies in Quadrant-IV.
The trigonometric ratios are as follows:
$\sin \theta =\dfrac{y}{r}=\dfrac{-15}{17} \\ \cos \theta =\dfrac{x}{r}=\dfrac{8}{17} \\ \tan \theta =\dfrac{y}{x}=\dfrac{-15}{8}$
and
$\csc \theta =\dfrac{r}{y}=\dfrac{-17}{15} \\ \sec \theta =\dfrac{r}{x}=\dfrac{17}{8} \\ \cot \theta =\dfrac{x}{y}=\dfrac{-8}{15}$
