cot^{-1}(-sqrt{3}) = | vedclass

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(B) We know that for the inverse cotangent function,the range is $(0, \pi)$.Since $\cot^{-1}(-x) = \pi - \cot^{-1}(x)$ for $x > 0$,we have:$\cot^{-1}(

Vedclass · Accepted Answer

$\frac{5\pi}{6}$

Vedclass · Answer

(B) We know that for the inverse cotangent function,the range is $(0, \pi)$.Since $\cot^{-1}(-x) = \pi - \cot^{-1}(x)$ for $x > 0$,we have:$\cot^{-1}(-\sqrt{3}) = \pi - \cot^{-1}(\sqrt{3})$We know that $\cot(\frac{\pi}{6}) = \sqrt{3}$,so $\cot^{-1}(\sqrt{3}) = \frac{\pi}{6}$.Therefore,$\cot^{-1}(-\sqrt{3}) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.

$\cot^{-1}(-\sqrt{3}) = $

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