Answer
$\sin \theta =\dfrac{y}{r}=\dfrac{\sqrt {10}}{10} \\ \cos \theta =\dfrac{x}{r}=\dfrac{- 3 \sqrt{10}}{10} \\ \tan \theta =\dfrac{y}{x}=\dfrac{-1}{3}$
and
$\csc \theta =\dfrac{r}{y}=\sqrt {10} \\ \sec \theta =\dfrac{r}{x}=-\dfrac{\sqrt {10}}{3} \\ \cot \theta =\dfrac{x}{y}=-3$
Work Step by Step
Here, $ x=-3; y=1$ and $ r=\sqrt {x^2+y^2}$
This gives: $ r=\sqrt {(-3)^2+(1)^2}=\sqrt {10} $; Because $\theta $ lies in Quadrant-II.
The trigonometric ratios are as follows:
$\sin \theta =\dfrac{y}{r}=\dfrac{\sqrt {10}}{10} \\ \cos \theta =\dfrac{x}{r}=\dfrac{- 3 \sqrt{10}}{10} \\ \tan \theta =\dfrac{y}{x}=\dfrac{-1}{3}$
and
$\csc \theta =\dfrac{r}{y}=\sqrt {10} \\ \sec \theta =\dfrac{r}{x}=-\dfrac{\sqrt {10}}{3} \\ \cot \theta =\dfrac{x}{y}=-3$
