Answer
$\sin \theta =\dfrac{y}{r}=\dfrac{-1}{4} \\ \cos \theta =\dfrac{x}{r}=\dfrac{-\sqrt {15}}{4} \\ \tan \theta =\dfrac{y}{x}=\dfrac{\sqrt {15}}{15}$
and
$\csc \theta =\dfrac{r}{y}=-4 \\ \sec \theta =\dfrac{r}{x}=-\dfrac{4 \sqrt {15}}{15} \\ \cot \theta =\dfrac{x}{y}=\sqrt {15}$
Work Step by Step
Here, $ y=-1; r=4$ and $ r=\sqrt {x^2+y^2}$
This gives: $ x=-\sqrt {(4)^2-(-1)^2}=- \sqrt {15}$; Because $\theta $ lies in Quadrant-III.
The trigonometric ratios are as follows:
$\sin \theta =\dfrac{y}{r}=\dfrac{-1}{4} \\ \cos \theta =\dfrac{x}{r}=\dfrac{-\sqrt {15}}{4} \\ \tan \theta =\dfrac{y}{x}=\dfrac{\sqrt {15}}{15}$
and
$\csc \theta =\dfrac{r}{y}=-4 \\ \sec \theta =\dfrac{r}{x}=-\dfrac{4 \sqrt {15}}{15} \\ \cot \theta =\dfrac{x}{y}=\sqrt {15}$
