The principal value of cot ^{-1}left(frac{-1} | vedclass

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Question

(A) We know that the range of the principal value branch of $\cot ^{-1} x$ is $(0, \pi)$.Since the argument $\frac{-1}{\sqrt{3}}$ is negative,we use t

Vedclass · Accepted Answer

$\frac{2 \pi}{3}$

Vedclass · Answer

(A) We know that the range of the principal value branch of $\cot ^{-1} x$ is $(0, \pi)$.Since the argument $\frac{-1}{\sqrt{3}}$ is negative,we use the property $\cot ^{-1}(-x) = \pi - \cot ^{-1}(x)$.Therefore,$\cot ^{-1}\left(\frac{-1}{\sqrt{3}}\right) = \pi - \cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)$.We know that $\cot \left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}}$,so $\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{3}$.Substituting this value,we get $\pi - \frac{\pi}{3} = \frac{2 \pi}{3}$.

The principal value of $\cot ^{-1}\left(\frac{-1}{\sqrt{3}}\right)$ is

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