Answer
$\sin \theta=\frac{-5}{13}$
$\cos \theta=\frac{-12}{13}$
$\tan \theta=\frac{5}{12}$
$\csc \theta=\frac{13}{-5}$
$\sec \theta=\frac{13}{-12}$
$\cot \theta=\frac{12}{5}$
Work Step by Step
We are given $x=-12, y=-5$
x,y,z form a right-angled triangle so to find z we can apply:
$x^2+y^2=z^2$
$\rightarrow z=\sqrt (-12)^2+(-5)^2=13$
With $x=-12, y=-5, z=13$, the trigonometric functions are
$\sin \theta=\frac{y}{z}=\frac{-5}{13}$
$\cos \theta=\frac{x}{z}=\frac{-12}{13}$
$\tan \theta=\frac{y}{x}=\frac{-5}{-12}=\frac{5}{12}$
$\csc \theta=\frac{1}{\sin \theta}=\frac{1}{\frac{-5}{13}}=\frac{13}{-5}$
$\sec \theta=\frac{1}{\cos \theta}=\frac{1}{\frac{-12}{13}}=\frac{13}{-12}$
$\cot \theta=\frac{x}{y}=\frac{12}{5}$
