Answer
$\sin\theta=\dfrac{\sqrt 5}{5}$
$\cos\theta=-\dfrac{2\sqrt 5}{5}$
$\tan\theta=-\dfrac{1}{2}$
$\sec\theta=-\dfrac{\sqrt 5}{2}$
$\csc\theta=\sqrt 5$
Work Step by Step
$\cot\theta=-2=\dfrac{x}{y}$
Since $\theta$ is in $\text{Quadrant II}$, we have:
$x=-2$
$y=1$
We find $r$ using the formula $r^2=x^2+y^2$,
$r^2=x^2+y^2$
$r^2=(-2)^2+1^2$
$r^2=5$
$r=\pm\sqrt5$
$r=\sqrt 5$ (since $r$ is never negative)
Compute for $\sin\theta, \cos\theta,\tan\theta, \sec\theta,\csc\theta$:
$\sin\theta=\dfrac{y}{r}=\dfrac{1}{\sqrt 5}=\dfrac{\sqrt 5}{5}$
$\cos\theta=\dfrac{x}{r}=\dfrac{-2}{\sqrt 5}=-\dfrac{2\sqrt 5}{5}$
$\tan\theta=\dfrac{y}{x}=\dfrac{1}{2}=-\dfrac{1}{2}$
$\sec\theta=\dfrac{r}{x}=\dfrac{\sqrt{5}}{-2}=-\dfrac{\sqrt 5}{2}$
$\csc\theta=\dfrac{r}{y}=\dfrac{\sqrt{5}}{1}=\sqrt 5$
