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($\frac {\pi}{6}$) = $\frac {\pi}{6}$