cos left[cot ^{-1}(-sqrt{3})+frac{pi}{6}right | vedclass

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(D) Given expression: $\cos \left[\cot ^{-1}(-\sqrt{3})+\frac{\pi}{6}\right]$ We know that $\cot ^{-1}(-x) = \pi - \cot ^{-1}(x)$. Therefore,$\cot ^{-

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$-1$

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(D) Given expression: $\cos \left[\cot ^{-1}(-\sqrt{3})+\frac{\pi}{6}\right]$ We know that $\cot ^{-1}(-x) = \pi - \cot ^{-1}(x)$. Therefore,$\cot ^{-1}(-\sqrt{3}) = \pi - \cot ^{-1}(\sqrt{3})$. Since $\cot \left(\frac{\pi}{6}\right) = \sqrt{3}$,we have $\cot ^{-1}(\sqrt{3}) = \frac{\pi}{6}$. Substituting this into the expression: $\cos \left[\pi - \frac{\pi}{6} + \frac{\pi}{6}\right]$ $= \cos [\pi]$ $= -1$.

$\cos \left[\cot ^{-1}(-\sqrt{3})+\frac{\pi}{6}\right]$ is equal to

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