Answer
$\sin{\theta} =-\dfrac{3}{5}$
$\tan{\theta}= \dfrac{3}{4}$
$\csc{\theta} = -\dfrac{5}{3}$
$\sec{\theta} =-\dfrac{5}{4}$
$\cot{\theta} =\dfrac{4}{3}$
Work Step by Step
Since $\theta \in QIII$, then $\sin{\theta} <0$.
Thus,
$\sin{\theta} = -\sqrt{1-\cos^2 {\theta}}$
$\sin{\theta} = -\sqrt{1-\left(-\dfrac{4}{5} \right)^2} = -\dfrac{3}{5}$
$\tan{\theta}= \dfrac{\sin{\theta}}{\cos{\theta}}$
$\tan{\theta}= \dfrac{-\dfrac{3}{5}}{-\dfrac{4}{5}} = \dfrac{3}{4}$
$\csc{\theta} = \dfrac{1}{\sin{\theta}}$
$\csc{\theta} = \dfrac{1}{-\dfrac{3}{5}}= -\dfrac{5}{3}$
$\sec{\theta} = \dfrac{1}{\cos{\theta}}$
$\sec{\theta} = \dfrac{1}{-\dfrac{4}{5}} = -\dfrac{5}{4}$
$\cot{\theta} = \dfrac{1}{\tan{\theta}}$
$\cot{\theta} = \dfrac{1}{\dfrac{3}{4}} = \dfrac{4}{3}$
