## Calculus with Applications (10th Edition)

$\sin \theta=\frac{-5}{13}$ $\cos \theta=\frac{-12}{13}$ $\tan \theta=\frac{5}{12}$ $\csc \theta=\frac{13}{-5}$ $\sec \theta=\frac{13}{-12}$ $\cot \theta=\frac{12}{5}$
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 $\sin \theta=\frac{y}{z}=\frac{-5}{13}$ $\cos \theta=\frac{x}{z}=\frac{-12}{13}$ $\tan \theta=\frac{y}{x}=\frac{-5}{-12}=\frac{5}{12}$ $\csc \theta=\frac{1}{\sin \theta}=\frac{1}{\frac{-5}{13}}=\frac{13}{-5}$ $\sec \theta=\frac{1}{\cos \theta}=\frac{1}{\frac{-12}{13}}=\frac{13}{-12}$ $\cot \theta=\frac{x}{y}=\frac{12}{5}$