Answer
$\sin{\theta} = \frac{-12}{13}
\\\cos{\theta} = \frac{-5}{13}
\\\cot{\theta}= \frac{5}{12}
\\\csc{\theta}= -\frac{13}{12}
\\\sec{\theta} = -\frac{13}{5}$
Work Step by Step
RECALL:
If P(x, y) is on the terminal side of an angle $\theta$ in standard position then:
$\begin{array}{cc}
&\sin{\theta} = \frac{y}{r} &\csc{\theta} = \frac{r}{y}
\\&\cos{\theta} = \frac{x}{r} &\sec{\theta} = \frac{r}{x}
\\&\tan{\theta} = \frac{y}{x} &\cot{\theta} = \frac{x}{y}
\end{array}$
where
$r= \sqrt{x^2+y^2}$
Theta is in Quadrant III where both x and y are negative.
Since $\tan{\theta} = \frac{y}{x}$, then $y=-12$ and $x=-5$.
Solve for $r$ to obtain:
$r=\sqrt{(-5)^2+(-12)^2} = \sqrt{25+144}=\sqrt{169}=13$
Thus, using the ratios in the recall part above, and with $P(-5, -12)$ and $r=13$, gives:
$\sin{\theta} = \frac{-12}{13}
\\\cos{\theta} = \frac{-5}{13}
\\\cot{\theta}=\frac{-5}{-12} = \frac{5}{12}
\\\csc{\theta}=\frac{13}{-12}=-\frac{13}{12}
\\\sec{\theta} = \frac{13}{-5} = -\frac{13}{5}$
