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

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Question

(B) Let $y = \cot^{-1}\left(\frac{-1}{\sqrt{3}}\right)$.Then $\cot y = \frac{-1}{\sqrt{3}}$.We know that the range of the principal value branch of $\

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $y = \cot^{-1}\left(\frac{-1}{\sqrt{3}}\right)$.Then $\cot y = \frac{-1}{\sqrt{3}}$.We know that the range of the principal value branch of $\cot^{-1} x$ is $(0, \pi)$.Since $\cot\left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}}$,we have $\cot\left(\pi - \frac{\pi}{3}\right) = -\cot\left(\frac{\pi}{3}\right) = \frac{-1}{\sqrt{3}}$.Thus,$\cot\left(\frac{2\pi}{3}\right) = \frac{-1}{\sqrt{3}}$.Since $\frac{2\pi}{3} \in (0, \pi)$,the principal value is $\frac{2\pi}{3}$.

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

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