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