Find the principal value of tan^{-1} \dfrac{1}{\sqrt{3}}

Given \( tan^{-1} \dfrac{1}{\sqrt{3}}\)

We know that for any x ∈ R, tan^-1 represents an angle in \( (\dfrac{-\pi}{2},\dfrac{\pi}{2})\) whose tangent is x.

So, \( tan^{-1} \dfrac{1}{\sqrt{3}}\) = an angle in \( (\dfrac{-\pi}{2},\dfrac{\pi}{2})\) whose tangent is \( \dfrac{1}{\sqrt{3}}\)

But we know that the value is equal to \( \dfrac{\pi}{6}\)

Therefore \( tan^{-1} \dfrac{1}{\sqrt{3}}=\dfrac{\pi}{6}\)

Hence the principal value of \( tan^{-1} \dfrac{1}{\sqrt{3}}=\dfrac{\pi}{6}\)