Answer
$\sin{\theta} =-\dfrac{3}{5}$
$\cos{\theta} =-\dfrac{4}{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
$ \tan{\theta} = \dfrac{y}{x} = \dfrac{3}{4}$
As $\theta$ terminates in $QIII$, any point on its terminal side will have negative $x$ and $y$ coordinates.
$\therefore y = -3 \hspace{20pt} x =-4$
$$r^2 = x^2+y^2 \\ r = \sqrt{x^2+y^2} = \sqrt{(-4)^2+(-3)^2} = 5$$
$\sin{\theta} = \dfrac{y}{r} = -\dfrac{3}{5}$
$\cos{\theta} = \dfrac{x}{r} = -\dfrac{4}{5} $
$\csc{\theta} = \dfrac{1}{\sin{\theta}} = -\dfrac{5}{3}$
$\sec{\theta} = \dfrac{1}{\cos{\theta}} = -\dfrac{5}{4}$
$\cot{\theta} = \dfrac{1}{\tan{\theta}} = \dfrac{4}{3}$
