cos ^{-1}left{cot left(sum_{i=1}^3 cot ^{-1} | vedclass

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(D) We need to evaluate the sum $S = \sum_{i=1}^3 \cot ^{-1} i = \cot ^{-1} 1 + \cot ^{-1} 2 + \cot ^{-1} 3$. Using the identity $\cot ^{-1} x = 	an

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$\frac{\pi}{2}$

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(D) We need to evaluate the sum $S = \sum_{i=1}^3 \cot ^{-1} i = \cot ^{-1} 1 + \cot ^{-1} 2 + \cot ^{-1} 3$. Using the identity $\cot ^{-1} x = 	an ^{-1} \frac{1}{x}$ for $x > 0$,we have $S = 	an ^{-1} 1 + 	an ^{-1} \frac{1}{2} + 	an ^{-1} \frac{1}{3}$. Since $	an ^{-1} 1 = \frac{\pi}{4}$,we calculate $	an ^{-1} \frac{1}{2} + 	an ^{-1} \frac{1}{3} = 	an ^{-1} \left( \frac{1/2 + 1/3}{1 - (1/2)(1/3)} ight) = 	an ^{-1} \left( \frac{5/6}{5/6} ight) = 	an ^{-1} 1 = \frac{\pi}{4}$. Thus,$S = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$. Now,we substitute this into the expression: $\cos ^{-1} \{ \cot (S) \} = \cos ^{-1} \{ \cot (\frac{\pi}{2}) \}$. Since $\cot (\frac{\pi}{2}) = 0$,the expression becomes $\cos ^{-1} (0)$. We know that $\cos ^{-1} (0) = \frac{\pi}{2}$. Therefore,the correct option is $D$.

$\cos ^{-1}\left\{\cot \left(\sum_{i=1}^3 \cot ^{-1} i\right)\right\}=$ . . . . . . .

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