The value of cot left(sum_{n=1}^{23} cot ^{-1 | vedclass

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(B) We know that $\sum_{k=1}^n 2k = 2 	imes \frac{n(n+1)}{2} = n(n+1)$.Substituting this into the expression,we get $\cot \left(\sum_{n=1}^{23} \cot

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$\frac{25}{23}$

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(B) We know that $\sum_{k=1}^n 2k = 2 	imes \frac{n(n+1)}{2} = n(n+1)$.Substituting this into the expression,we get $\cot \left(\sum_{n=1}^{23} \cot ^{-1}(1+n(n+1))ight)$.Using the identity $\cot^{-1}(x) = 	an^{-1}(\frac{1}{x})$ for $x > 0$,the expression becomes $\cot \left(\sum_{n=1}^{23} 	an ^{-1}\left(\frac{1}{1+n(n+1)}ight)ight)$.We can rewrite the argument of $	an^{-1}$ as $\frac{(n+1)-n}{1+n(n+1)}$,which allows us to use the formula $	an^{-1}(a) - 	an^{-1}(b) = 	an^{-1}(\frac{a-b}{1+ab})$.Thus,$\sum_{n=1}^{23} 	an ^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}ight) = \sum_{n=1}^{23} (	an^{-1}(n+1) - 	an^{-1}(n))$.This is a telescoping sum: $(	an^{-1}(2) - 	an^{-1}(1)) + (	an^{-1}(3) - 	an^{-1}(2)) + \dots + (	an^{-1}(24) - 	an^{-1}(23)) = 	an^{-1}(24) - 	an^{-1}(1)$.Now,$\cot(	an^{-1}(24) - 	an^{-1}(1)) = \cot(	an^{-1}(\frac{24-1}{1+24 	imes 1})) = \cot(	an^{-1}(\frac{23}{25}))$.Since $\cot(	an^{-1}(x)) = \cot(\cot^{-1}(\frac{1}{x})) = \frac{1}{x}$,we have $\cot(	an^{-1}(\frac{23}{25})) = \frac{25}{23}$.

The value of $\cot \left(\sum_{n=1}^{23} \cot ^{-1}\left(1+\sum_{k=1}^n 2 k\right)\right)$ is

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