cot^{-1}(1) + cot^{-1} (frac{1}{2}) + cot^{-1 | vedclass

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(D) We know that $\cot^{-1}(x) = 	an^{-1}(\frac{1}{x})$ for $x > 0$.Therefore,the expression becomes:$	an^{-1}(1) + 	an^{-1}(2) + 	an^{-1}(3)$We k

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$\pi$

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(D) We know that $\cot^{-1}(x) = 	an^{-1}(\frac{1}{x})$ for $x > 0$.Therefore,the expression becomes:$	an^{-1}(1) + 	an^{-1}(2) + 	an^{-1}(3)$We know that $	an^{-1}(1) = \frac{\pi}{4}$.Now,for $	an^{-1}(2) + 	an^{-1}(3)$,since $2 	imes 3 = 6 > 1$,we use the formula $	an^{-1}(x) + 	an^{-1}(y) = \pi + 	an^{-1}(\frac{x+y}{1-xy})$:$	an^{-1}(2) + 	an^{-1}(3) = \pi + 	an^{-1}(\frac{2+3}{1-2 	imes 3})$$= \pi + 	an^{-1}(\frac{5}{1-6})$$= \pi + 	an^{-1}(\frac{5}{-5})$$= \pi + 	an^{-1}(-1)$$= \pi - 	an^{-1}(1)$$= \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.Adding these results together:$\frac{\pi}{4} + \frac{3\pi}{4} = \frac{4\pi}{4} = \pi$.

$\cot^{-1}(1) + \cot^{-1} (\frac{1}{2}) + \cot^{-1}(\frac{1}{3}) =$

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