Answer
$\sin{\theta} =\dfrac{\sqrt{5}}{5}$
$\cos{\theta} =-\dfrac{2\sqrt{5}}{5} $
$\tan{\theta} =-\dfrac{1}{2}$
$\csc{\theta} =\sqrt{5}$
$\sec{\theta} =-\dfrac{\sqrt{5}}{2}$
$ \cot{\theta} = -2$
Work Step by Step
$ \cot{\theta} = \dfrac{x}{y} = -\dfrac{2}{1}$
As $\theta$ terminates in $QII$, any point on its terminal side will have a negative $x$ coordinate and a positive $y$ coordinate.
$\therefore x = -2 \hspace{20pt} y =1$
$$r^2 = x^2+y^2 \\ r = \sqrt{x^2+y^2} = \sqrt{(-2)^2+(1)^2} = \sqrt{5}$$
$\sin{\theta} = \dfrac{y}{r} = \dfrac{\sqrt{5}}{5}$
$\cos{\theta} = \dfrac{x}{r} = -\dfrac{2\sqrt{5}}{5} $
$\tan{\theta} = \dfrac{y}{x} = -\dfrac{1}{2}$
$\csc{\theta} = \dfrac{1}{\sin{\theta}} = \sqrt{5}$
$\sec{\theta} = \dfrac{1}{\cos{\theta}} = -\dfrac{\sqrt{5}}{2}$
