Answer
$\sin\theta=\dfrac{3}{5}$
$\cos\theta=-\dfrac{4}{5}$
$\tan\theta=-\dfrac{3}{4}$
$\cot\theta=-\dfrac{4}{3}$
$\csc\theta=\dfrac{5}{3}$
Work Step by Step
First we determine $\cos\theta$:
$\cos\theta=\dfrac{1}{\sec\theta}=\dfrac{1}{-\dfrac{5}{4}}=-\dfrac{4}{5}$
Since $\tan\theta<0$ and $\cos\theta<0$, then angle $\theta$ is in $\text{Quadrant II}$.
We have:
$\cos\theta=\dfrac{x}{r}=-\dfrac{4}{5}$
Thus,
$x=-4$
$r=5$
We find $y$ by using the formula $r^2=x^2+y^2$:
$5^2=(-4)^2+y^2$
$25=16+y^2$
$9=y^2$
$\pm\sqrt9=y$
$\pm3=y$
Since $\theta$ is in $\text{Quadrant II}$ where $y$ is positive, then $y=3$.
Compute $\sin\theta, \tan\theta, \cot\theta,\csc\theta$:
$\sin\theta=\dfrac{y}{r}=\dfrac{3}{5}$
$\tan\theta=\dfrac{y}{x}=\dfrac{3}{-4}=-\dfrac{3}{4}$
$\cot\theta=\dfrac{x}{y}=\dfrac{-4}{3}=-\dfrac{4}{3}$
$\csc\theta=\dfrac{r}{y}=\dfrac{5}{3}$
