Krishna
0

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 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