Answer
$sin(\theta) =-\frac{3\sqrt {13}}{13}$,
$cos(\theta) =\frac{2\sqrt {13}}{13}$,
$tan(\theta) =-\frac{3}{2}$,
$cot(\theta) =-\frac{2}{3}$,
$sec(\theta) =\frac{\sqrt {13}}{2}$,
$csc(\theta) =-\frac{\sqrt {13}}{3}$.
Work Step by Step
Given $x=2,y=-3$, the terminal side is in quadrant IV with $r=\sqrt {x^2+y^2}=\sqrt {13}$, we have:
$sin(\theta)=\frac{y}{r}=-\frac{3\sqrt {13}}{13}$,
$cos(\theta)=\frac{x}{r}=\frac{2\sqrt {13}}{13}$,
$tan(\theta)=\frac{y}{x}=-\frac{3}{2}$,
$cot(\theta)=\frac{x}{y}=-\frac{2}{3}$,
$sec(\theta)=\frac{r}{x}=\frac{\sqrt {13}}{2}$,
$csc(\theta)=\frac{r}{y}=-\frac{\sqrt {13}}{3}$.
