Answer
$\sin{\theta} =\dfrac{\sqrt{2}}{2}$
$\cos{\theta} =\dfrac{\sqrt{2}}{2}$
$\tan{\theta} = 1$
$\csc{\theta} =\sqrt{2}$
$\sec{\theta} =\sqrt{2}$
$\cot{\theta} =1$
Work Step by Step
$\because \cos{\theta} = \dfrac{x}{r} = \dfrac{\sqrt{2}}{2}$
$\therefore x = \sqrt{2} \hspace{10pt} \& \hspace{10pt} r = 2$
$$r^2 = x^2+y^2 \\ y^2 = r^2-x^2 \\ y = \pm \sqrt{r^2-x^2} =\pm \sqrt{2^2-(\sqrt{2})^2} = \pm \sqrt{2}$$
As $\theta$ terminates in $QI$, any point on its terminal side will have a positive $y$ coordinate.
$\therefore y = \sqrt{2}$
$\sin{\theta} = \dfrac{y}{r} = \dfrac{\sqrt{2}}{2}$
$\tan{\theta} = \dfrac{y}{x} = \dfrac{\sqrt{2}}{\sqrt{2}} = 1$
$\csc{\theta} = \dfrac{1}{\sin{\theta}} = \sqrt{2}$
$\sec{\theta} = \dfrac{1}{\cos{\theta}} = \sqrt{2}$
$\cot{\theta} = \dfrac{1}{\tan{\theta}} = 1$
