Answer
The exact value of the expression is $1$.
Work Step by Step
The $\tan \left( -\frac{11\pi }{4} \right)$ lies in quadrant III.
Add $4\pi $ to $-\frac{11\pi }{4}$, to find a positive co-terminal angle less than $2\pi $.
$\begin{align}
& \alpha =\left( -\frac{11\pi }{4} \right)+4\pi \\
& =\frac{-11\pi +16\pi }{4} \\
& =\frac{5\pi }{4}
\end{align}$
The positive acute angle formed by the terminal side of $\theta $ and the x-axis is the reference angle ${\theta }'$.
Since, $\frac{5\pi }{4}$ angle lies in quadrant III, subtract $\pi $ from $\frac{5\pi }{4}$ to find the reference angle.
$\begin{align}
& \theta '=\frac{5\pi }{4}-\pi \\
& =\frac{5\pi -\pi }{4} \\
& =\frac{\pi }{4}
\end{align}$
The reference angle of $-\frac{11\pi }{4}$ is $\frac{\pi }{4}$.
The function value of the reference angle is
$\tan \frac{\pi }{4}=1$
The angle $-\frac{11\pi }{4}$ is in quadrant III and the tangent function is positive in quadrant III.
Therefore,
$\tan \left( -\frac{11\pi }{4} \right)=\tan \frac{\pi }{4}$
Substitute $1$ for $\tan \frac{\pi }{4}$.
$\tan \left( -\frac{11\pi }{4} \right)=1$
The exact value of the expression is $1$.