Answer
$a.\displaystyle \quad\frac{\pi}{4}$
$b.\displaystyle \quad-\frac{\pi}{3}$
$c.\displaystyle \quad\frac{\pi}{6}$
Work Step by Step
$y=\tan^{-1}x$ is the number in $(-\pi/2, \pi/2)$ for which $\tan y=x.$
Use reference angles in the 1st quadrant,
keeping in mind that tan is an odd function.
$\left\{\begin{array}{llll}
\tan\frac{\pi}{4}=1 & & & \Rightarrow\tan^{-1}1=\dfrac{\pi}{4}\\
\tan\frac{\pi}{3}=\sqrt{3} & \Rightarrow & \tan(-\frac{\pi}{3})=-\sqrt{3} & \Rightarrow\tan^{-1}(-\sqrt{3})=-\dfrac{\pi}{3}\\
\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}} & & & \Rightarrow\tan^{-1}(\frac{1}{\sqrt{3}})=\dfrac{\pi}{6}
\end{array}\right.$
