Answer
$ \sin{t} = -\dfrac{1}{3}$
$ \cos{t} = \dfrac{2\sqrt{2}}{3}$
$\tan{t} = -\dfrac{\sqrt{2}}{4}$
$\csc{t} =-3$
$\sec{t} =\dfrac{3\sqrt{2}}{4}$
$\cot{t} =-2 \sqrt{2}$
Work Step by Step
With $P= \left(\dfrac{2\sqrt{2}}{3},-\dfrac{1}{3} \right) = (x,y)$, then $\hspace{15pt}x = \dfrac{2\sqrt{2}}{3} \text{ and } \hspace{10pt} y =-\dfrac{1}{3}$
Thus,
$\sin{t} = y$
$ \sin{t} = -\dfrac{1}{3}$
$\cos{t} = x$
$ \cos{t} = \dfrac{2\sqrt{2}}{3}$
$\tan{t} = \dfrac{y}{x}$
$\tan{t} = \dfrac{-\dfrac{1}{3}}{ \dfrac{2\sqrt{2}}{3} } = -\dfrac{\sqrt{2}}{4}$
$\csc{t} = \dfrac{1}{y}$
$\csc{t} = \dfrac{1}{-\dfrac{1}{3}} = -3$
$\sec{t} = \dfrac{1}{x}$
$\sec{t} = \dfrac{1}{\dfrac{2\sqrt{2}}{3}}= \dfrac{3\sqrt{2}}{4}$
$\cot{t} = \dfrac{x}{y}$
$\cot{t} = \dfrac{\dfrac{2\sqrt{2}}{3} }{-\dfrac{1}{3}} = -2 \sqrt{2}$
