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

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(C) We know that $	an^{-1}(\sqrt{3}) = \frac{\pi}{3}$.Also,we use the property $\cot^{-1}(-x) = \pi - \cot^{-1}(x)$.Therefore,$\cot^{-1}(-\sqrt{3}) =

Vedclass · Accepted Answer

$-\frac{\pi}{2}$

Vedclass · Answer

(C) We know that $	an^{-1}(\sqrt{3}) = \frac{\pi}{3}$.Also,we use the property $\cot^{-1}(-x) = \pi - \cot^{-1}(x)$.Therefore,$\cot^{-1}(-\sqrt{3}) = \pi - \cot^{-1}(\sqrt{3})$.Since $\cot^{-1}(\sqrt{3}) = \frac{\pi}{6}$,we have $\cot^{-1}(-\sqrt{3}) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.Now,substituting these values into the expression:$	an^{-1}(\sqrt{3}) - \cot^{-1}(-\sqrt{3}) = \frac{\pi}{3} - \frac{5\pi}{6}$.Finding a common denominator:$\frac{2\pi}{6} - \frac{5\pi}{6} = -\frac{3\pi}{6} = -\frac{\pi}{2}$.Thus,the correct option is $C$.

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

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