Answer
$\sin\theta=\frac{y}{r}=\frac{-3}{5}$
$\cos\theta=\frac{x}{r}=\frac{4}{5}$
$\csc\theta=\frac{r}{y}=\frac{5}{-3}$
$\sec\theta=\frac{r}{x}=\frac{5}{4}$
$\tan\theta=\frac{y}{x}=\frac{-3}{4}$
$\cot\theta=\frac{x}{y}=\frac{4}{-3}$
Work Step by Step
From secant and quadrant we know that $x=4; r=5$, so lets find x:
$x^{2}+y^{2}=r^{2}$
$4^{2}+y^{2}=5^{2}$
$y^{2}+16=25$
$y^{2}=9$ (since $y\lt0$ in IV quadrant)
$y=-3$
