Find the exact value of \tan^{-1}(- \sqrt{3})

Step 1: Find the given inverse trigonometric function
GIVEN: \tan^{-1}(-\sqrt{3})
Let \tan^{-1}(-\sqrt{3})=\theta
\tan\theta=-\sqrt{3}
A basic reference triangle looks like:
\tan \theta = \frac{opposite}{adjacent}
\tan 60\degree = \frac{\sqrt{3}}{1}
Step 2: Locate the negative quadrants of \tan
\tan \theta = -\sqrt{3}
Tan values are negative in the second and fourth quadrants.
So, Second quadrant \theta = 180\degree - 60\degree
\theta = 120\degree
Or
Fourth quadrant \theta = 3600\degree - 60\degree
\theta = 300\degree
120 which is a rotation through 180°
\tan x=n\ if\ and\ only\ if\ x=\arctan n+\pi k for some integer k and so on.
Hence, \theta = 120\degree + \pi k