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

Asked by Sakshi | 1 year ago |  80

1 Answer

Solution :-

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

Answered by Aaryan | 1 year ago

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