Answer
$\sin{t} =- \dfrac{\sqrt{17}}{17}$
$\cos{t} = -\dfrac{4\sqrt{17}}{17}$
$\csc{t} = -\sqrt{17}$
$\sec{t} =-\dfrac{\sqrt{17}}{4}$
$\cot{t} =4$
Work Step by Step
$\tan{t} =\dfrac{1}{4}$
$\because t$ terminates in $ QIII \hspace{20pt} \therefore \sec{t} $ is negative
$\sec^2{t} = 1+\tan^2{t} \\ \sec{t} = -\sqrt{1+\tan^2{t}} \\ = -\sqrt{1+\left(\dfrac{1}{4}\right)^2} = \sec{t} =-\dfrac{\sqrt{17}}{4}$
$\cos{t} = \dfrac{1}{\sec{t}} = -\dfrac{4\sqrt{17}}{17}$
$\sin{t} = \cos{t} \times \tan{t} = - \dfrac{\sqrt{17}}{17}$
$\csc{t} = \dfrac{1}{\sin{t}} = -\sqrt{17}$
$\cot{t} = \dfrac{1}{\tan{t}} =4$
