Answer
$\sin{t} =-\dfrac{2\sqrt{2}}{3}$
$\cos{t} = \dfrac{1}{3}$
$\tan{t} =-2\sqrt{2}$
$\csc{t} =-\dfrac{3\sqrt{2}}{4}$
$\cot{t} = -\dfrac{\sqrt{2}}{4}$
Work Step by Step
$\because\sec{t} =3 \hspace{20pt} \therefore \cos{t} = \dfrac{1}{3}$
$\because t$ terminates in $ QIV \hspace{20pt} \therefore \sin{t} $ is negative
$\sin^2{t} = 1-\cos^2{t} \\ \sin{t} = -\sqrt{1-\cos^2{t}} \\ = -\sqrt{1-\left(\dfrac{1}{3}\right)^2} = -\dfrac{2\sqrt{2}}{3}$
$\tan{t} = \dfrac{\sin{t}}{\cos{t}} = -2\sqrt{2}$
$\csc{t} =\dfrac{1}{\sin{t}} = -\dfrac{3\sqrt{2}}{4}$
$\cot{t} = \dfrac{1}{\tan{t}}-\dfrac{\sqrt{2}}{4}$