The value of cot ^{-1}(-sqrt{3})-tan ^{-1} sq | vedclass

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Question

(C) We know that $\cot ^{-1}(-x) = \pi - \cot ^{-1}(x)$ for $x > 0$. Given expression is $\cot ^{-1}(-\sqrt{3}) - 	an ^{-1}(\sqrt{3})$. Using the pro

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$\frac{\pi}{2}$

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(C) We know that $\cot ^{-1}(-x) = \pi - \cot ^{-1}(x)$ for $x > 0$. Given expression is $\cot ^{-1}(-\sqrt{3}) - 	an ^{-1}(\sqrt{3})$. Using the property,$\cot ^{-1}(-\sqrt{3}) = \pi - \cot ^{-1}(\sqrt{3})$. Since $\cot ^{-1}(\sqrt{3}) = \frac{\pi}{6}$ and $	an ^{-1}(\sqrt{3}) = \frac{\pi}{3}$,Substituting these values: $\cot ^{-1}(-\sqrt{3}) - 	an ^{-1}(\sqrt{3}) = (\pi - \frac{\pi}{6}) - \frac{\pi}{3}$. $= \frac{5\pi}{6} - \frac{\pi}{3} = \frac{5\pi - 2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}$. Thus,the correct option is $C$.

The value of $\cot ^{-1}(-\sqrt{3})-\tan ^{-1} \sqrt{3}$ is equal to . . . . . . .

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