Question

Find the value of \(tan^{-1}(tan \frac{7\pi}{6})\).

Answer 1

We know that tan⁻¹(tan x) = x if x∈\([-\frac {\pi}{2}, \frac {\pi}{2}]\), which is the principal value branch of tan⁻¹x. 
Here,\(\frac {7\pi}{6}\) ∉ \([-\frac {\pi}{2}, \frac {\pi}{2}]\). 
Now,tan^-1(tan\(\frac {7\pi}{6}\)) can be written as: 
tan^-1(tan\(\frac {7\pi}{6}\)) = tan^-1(tan 2\(\pi\)-\(\frac {5\pi}{6}\))     [tan(2\(\pi\)-x) = -tan x] 
=tan^-1(-tan\(\frac {5\pi}{6}\)) = tan^-1(-tan\(\frac {5\pi}{6}\)) = tan^-1(tan(-\(\frac {5\pi}{6}\))) = tan^-1(tan(\(\pi\)-\(\frac {5\pi}{6}\))) 
= tan^-1(tan(\(\frac {\pi}{6}\))), where \(\frac {\pi}{6}\) ∈ \([-\frac {\pi}{2}, \frac {\pi}{2}]\) 

Therefore tan^-1(tan\(\frac {7\pi}{6}\)) = tan^-1(tan(\(\frac {\pi}{6}\)) = \(\frac {\pi}{6}\)