Answer
$\sin\theta=\frac{y}{r}=\frac{2}{\sqrt 5}=\frac{2\sqrt 5}{5}$
$\cos\theta=\frac{x}{r}=\frac{-1}{\sqrt 5}=-\frac{\sqrt 5}{5}$
$\csc\theta=\frac{r}{y}=\frac{\sqrt 5}{2}$
$\sec\theta=\frac{r}{x}=\frac{\sqrt 5}{-1}=-\sqrt 5$
$\tan\theta=\frac{y}{x}=\frac{2}{-1}=-2$
$\cot\theta=\frac{x}{y}=\frac{-1}{2}$
Work Step by Step
From secant and II quadrant we know that $x=-1; r=\sqrt 5$, so lets find x:
$x^{2}+y^{2}=r^{2}$
$(-1)^{2}+y^{2}=(\sqrt 5)^{2}$
$y^{2}+1=25$
$y^{2}=4$ (since in II quadrant y is positive)
$y=2$
