Answer
The six trigonometric functions of $\theta $ for point $\left( -1,-3 \right)$ are,
$\sin \theta =-\frac{3\sqrt{10}}{\sqrt{10}},\cos \theta =-\frac{\sqrt{10}}{10},\tan \theta =3,\csc \theta =-\frac{\sqrt{10}}{3},\sec \theta =-\sqrt{10}$ and $\cot \theta =\frac{1}{3}$.
Work Step by Step
Consider the point $\left( -1,-3 \right)$.
Here, $x=-1$ and $y=-3$.
The six trigonometric functions of $\theta $ are defined in terms of a ratio by using the right-angle triangle.
In the figure, $y$ is the perpendicular, $x$ is the base and $r$ is the hypotenuse.
According to the Pythagoras theorem, the hypotenuse is,
$r=\sqrt{{{x}^{2}}+{{y}^{2}}}$
Substitute $-1$ for $x$ and $-3$ for $y$.
$\begin{align}
& r=\sqrt{{{\left( -1 \right)}^{2}}+{{\left( -3 \right)}^{2}}} \\
& =\sqrt{1+9} \\
& =\sqrt{10}
\end{align}$
Recall the trigonometric expression of $\sin \theta $.
$\sin \theta =\frac{y}{r}$
Substitute $-3$ for $y$ and $\sqrt{10}$ for $r$.
$\begin{align}
& \sin \theta =-\frac{3}{\sqrt{10}} \\
& =-\frac{3}{\sqrt{10}}\cdot \frac{\sqrt{10}}{\sqrt{10}} \\
& =-\frac{3\sqrt{10}}{10}
\end{align}$
Recall the trigonometric expression of $\cos \theta $.
$\cos \theta =\frac{x}{r}$
Substitute $-1$ for $x$ and $\sqrt{10}$ for $r$.
$\begin{align}
& \cos \theta =-\frac{1}{\sqrt{10}} \\
& =-\frac{1}{\sqrt{10}}\cdot \frac{\sqrt{10}}{\sqrt{10}} \\
& =-\frac{\sqrt{10}}{10}
\end{align}$
Recall the trigonometric expression of $\tan \theta $.
$\tan \theta =\frac{y}{x}$
Substitute $-1$ for $x$ and $-3$ for $y$.
$\begin{align}
& \tan \theta =\frac{-3}{-1} \\
& =3
\end{align}$
Recall the trigonometric expression of $\csc \theta $.
$\csc \theta =\frac{r}{y}$
Substitute $-3$ for $y$ and $\sqrt{10}$ for $r$.
$\csc \theta =-\frac{\sqrt{10}}{3}$
Recall the trigonometric expression of $\sec \theta $.
$\sec \theta =\frac{r}{x}$
Substitute $-1$ for $x$ and $\sqrt{10}$ for $r$.
$\begin{align}
& \sec \theta =-\frac{\sqrt{10}}{1} \\
& =-\sqrt{10}
\end{align}$
Recall the trigonometric expression of $\cot \theta $.
$\cot \theta =\frac{x}{y}$
Substitute $-1$ for $x$ and $-3$ for $y$.
$\begin{align}
& \cot \theta =\frac{-1}{-3} \\
& =\frac{1}{3}
\end{align}$
Thus, the six trigonometric functions of $\theta $ for point $\left( -1,-3 \right)$ are,
$\sin \theta =-\frac{3\sqrt{10}}{\sqrt{10}},\cos \theta =-\frac{\sqrt{10}}{10},\tan \theta =3,\csc \theta =-\frac{\sqrt{10}}{3},\sec \theta =-\sqrt{10}$ and $\cot \theta =\frac{1}{3}$.
