Answer
$-\dfrac{\pi}{4}$
Work Step by Step
$y = \tan^{-1}{x} \hspace{15pt} \to \hspace{15pt} x = \tan{y}$
$\text{where } \hspace{15pt} -\infty \leq x \leq \infty \hspace{15pt} \text{and} \hspace{15pt} -\dfrac{\pi}{2} \leq y \leq \dfrac{\pi}{2}$
Let $\theta = \tan^{-1} {(-1)} \hspace{20pt} -\dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2}$
Since $\theta = \tan^{-1}{(-1)}$, then
$\tan{\theta} =-1$
Referring to Table 3 on page 472:
The only angle in the interval $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2} \right]$ whose tangent is $-1$ is $-\dfrac{\pi}{4}$.
Therefore, $\tan^{-1} {(-1)} = -\dfrac{\pi}{4}$.
