Answer
The trigonometric functions are:
$\sin \theta =-\frac{\sqrt{10}}{10}\text{; cos}\theta \text{=}-\frac{3\sqrt{10}}{10}\text{; tan}\theta \text{=}\frac{1}{3};\text{ csc}\theta \text{=}-\sqrt{10};\text{ sec}\theta \text{=}-\frac{\sqrt{10}}{3}$.
Work Step by Step
We know that $\cot \theta >0\text{ and cos}\theta <0$ , so $\theta $ lies in quadrant III. So, in quadrant III, x and y are both negative. Here,
$\begin{align}
& \cot \theta =\frac{3}{1} \\
& =\frac{x}{y} \\
& =\frac{-3}{-1}
\end{align}$
Therefore, $x=-3\text{ and }y=-1$. Now, calculate r as:
$\begin{align}
& {{r}^{2}}={{x}^{2}}+{{y}^{2}} \\
& r=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( -1 \right)}^{2}}} \\
& r=\sqrt{9+1} \\
& r=\sqrt{10}
\end{align}$
Now, evaluate the sine function as follows:
$\begin{align}
& \sin \theta =\frac{y}{r} \\
& =\frac{-1}{\sqrt{10}} \\
& =\frac{-1\cdot \sqrt{10}}{\sqrt{10}\cdot \sqrt{10}} \\
& =-\frac{\sqrt{10}}{10}
\end{align}$
Then, 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 tangent function as follows:
$\begin{align}
& \tan \theta =\frac{y}{x} \\
& =\frac{-1}{-3} \\
& =\frac{1}{3}
\end{align}$
Then, evaluate the cosecant function as follows:
$\begin{align}
& \csc \theta =\frac{r}{y} \\
& =\frac{\sqrt{10}}{-1} \\
& =-\sqrt{10}
\end{align}$
Finally evaluate the secant function as follows:
$\begin{align}
& \sec \theta =\frac{r}{x} \\
& =\frac{\sqrt{10}}{-3} \\
& =-\frac{\sqrt{10}}{3}
\end{align}$
