Evaluate tan-1(tan 7π/6)

Given \( tan^{-1}(tan \dfrac{7\pi}{6})\)

We know that tan  \(tan \dfrac{7\pi}{6} =\dfrac{1}{\sqrt{3}}\)

By substituting this value in \( tan^{-1}(tan \dfrac{7\pi}{6})\)we get,

Tan^-1 \( \dfrac{1}{\sqrt{3}}\)

Now let tan^-1 \(( \dfrac{1}{\sqrt{3}})=y\)

Tan y = \( \dfrac{1}{\sqrt{3}}\)

 \(tan \dfrac{7\pi}{6} =\dfrac{1}{\sqrt{3}} \)

The range of the principal value of tan^-1 is \( (\dfrac{-\pi}{2},\dfrac{\pi}{2})\)

and  \(tan \dfrac{7\pi}{6} =\dfrac{1}{\sqrt{3}} \)

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