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