Answer
$sin\theta= -\frac{\sqrt {15}}{4}$,
$tan\theta= \sqrt {15}$,
$cot\theta= \frac{\sqrt {15}}{15}$,
$sec\theta= -4$,
$csc\theta= -\frac{4\sqrt {15}}{15}$.
Work Step by Step
Given $cos\theta=-\frac{1}{4}$, let $x=1, r=4$, we can get $y=\sqrt {r^2-x^2}=\sqrt {15}$. Use the fact that $tan\theta\gt0,cos\theta\lt0\longrightarrow \theta$ is in quadrant III to determine the sign of each function:
$sin\theta=-\frac{y}{r}=-\frac{\sqrt {15}}{4}$,
$tan\theta=\frac{y}{x}=\sqrt {15}$,
$cot\theta=\frac{x}{y}=\frac{\sqrt {15}}{15}$,
$sec\theta=-\frac{r}{x}=-4$,
$csc\theta=-\frac{r}{y}=-\frac{4\sqrt {15}}{15}$.
