Answer
The trigonometric functions are:
$\sin \theta =\frac{\sqrt{10}}{10}\text{; cos}\theta \text{=}-\frac{3\sqrt{10}}{10}\text{; csc}\theta \text{=}\sqrt{10};\text{ sec}\theta \text{=}-\frac{\sqrt{10}}{3};\text{ cot}\theta \text{=}-3$.
Work Step by Step
Since $\tan \theta <0\text{ and sin}\theta >0$, so $\theta $ lies in quadrant II. In quadrant II, x is negative and y is positive. Here,
$\begin{align}
& \tan \theta =-\frac{1}{3} \\
& =\frac{y}{x}
\end{align}$
So, $ x=-3\text{ and }y=1$. Now, calculate r as:
$\begin{align}
& {{r}^{2}}={{x}^{2}}+{{y}^{2}} \\
& r=\sqrt{{{\left( -3 \right)}^{2}}+{{1}^{2}}} \\
& r=\sqrt{10}
\end{align}$
Then, evaluate the sine function as follows:
$\begin{align}
& \sin \theta =\frac{y}{r} \\
& =\frac{1}{\sqrt{10}} \\
& =\frac{1\cdot \sqrt{10}}{\sqrt{10}\cdot \sqrt{10}} \\
& =\frac{\sqrt{10}}{10}
\end{align}$
Now, evaluate the cosine function as follows:
$\begin{align}
& \cos \theta =\frac{x}{r} \\
& =\frac{-3}{\sqrt{10}} \\
& =-\frac{3\cdot \sqrt{10}}{\sqrt{10}\cdot \sqrt{10}} \\
& =-\frac{3\sqrt{10}}{10}
\end{align}$
Then, evaluate the cosecant function as follows:
$\begin{align}
& \csc \theta =\frac{r}{y} \\
& =\frac{\sqrt{10}}{1} \\
& =\sqrt{10}
\end{align}$
Then, evaluate the secant function as follows:
$\begin{align}
& \sec \theta =\frac{r}{x} \\
& =\frac{\sqrt{10}}{-3} \\
& =-\frac{\sqrt{10}}{3}
\end{align}$
Also, evaluate the cotangent function as follows:
$\begin{align}
& \cot \theta =\frac{x}{y} \\
& =\frac{-3}{1} \\
& =-3
\end{align}$
