Answer
$\sin{\theta}= -\dfrac{1}{2}$
$\cos{\theta} =\dfrac{\sqrt{3}}{2}$
$\tan{\theta} =-\dfrac{\sqrt{3}}{3}$
$\csc{\theta} =-2$
$\sec{\theta} = \dfrac{2\sqrt{3}}{3}$
$\cot{\theta} = -\sqrt{3}$
Work Step by Step
$\because \cos{\theta} = \dfrac{x}{r} = \dfrac{\sqrt{3}}{2}$
$\therefore x = \sqrt{3} \hspace{10pt} \& \hspace{10pt} r = 2$
$$r^2 = x^2+y^2 \\ y^2 = r^2-y^2 \\ y = \pm \sqrt{r^2-y^2} =\pm \sqrt{2^2-(\sqrt{3})^2} = \pm 1$$
As $\theta$ terminates in $QIV$, any point on its terminal side will have a negative $y$ coordinate.
$\therefore y = -1$
$\sin{\theta} = \dfrac{y}{r} = -\dfrac{1}{2}$
$\tan{\theta} = \dfrac{y}{x} = -\dfrac{\sqrt{3}}{3}$
$\csc{\theta} = \dfrac{1}{\sin{\theta}} = -2$
$\sec{\theta} = \dfrac{1}{\cos{\theta}} = \dfrac{2\sqrt{3}}{3}$
$\cot{\theta} = \dfrac{1}{\tan{\theta}} = -\sqrt{3}$
