Answer
Here, $ x= -4; y=3$ $ r=\sqrt {x^2+y^2}=\sqrt {(-4)^2+(3)^2}=5$
The trigonometric ratios are as follows:
$ sin \theta =\dfrac{y}{r}=\dfrac{3}{5} \\cos \theta =\dfrac{x}{r}=\dfrac{-4}{5} \\ \tan \theta =\dfrac{y}{x}=\dfrac{-3}{4}$
and
$\csc \theta =\dfrac{r}{y}=\dfrac{5}{3} \\ \sec \theta =\dfrac{r}{x}=\dfrac{-5}{4} \\ \cot \theta =\dfrac{x}{y}=\dfrac{-4}{3}$
Work Step by Step
Here, $ x= -12; y=5$ $ r=\sqrt {x^2+y^2}=\sqrt {(-12)^2+(5)^2}=13$
The trigonometric ratios are as follows:
$ sin \theta =\dfrac{y}{r}=\dfrac{5}{13} \\cos \theta =\dfrac{x}{r}=\dfrac{-12}{13} \\ \tan \theta =\dfrac{y}{x}=\dfrac{-5}{12}$
and
$\csc \theta =\dfrac{r}{y}=\dfrac{13}{5} \\ \sec \theta =\dfrac{r}{x}=\dfrac{-13}{12} \\ \cot \theta =\dfrac{x}{y}=\dfrac{-12}{5}$
