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

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(B) We know that $\sum_{k=1}^n 2k = 2 	imes \frac{n(n+1)}{2} = n(n+1)$.So,the expression inside the summation becomes $\cot^{-1}(1 + n(n+1))$.Using t

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$\frac{25}{23}$

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(B) We know that $\sum_{k=1}^n 2k = 2 	imes \frac{n(n+1)}{2} = n(n+1)$.So,the expression inside the summation becomes $\cot^{-1}(1 + n(n+1))$.Using the identity $\cot^{-1}(x) = 	an^{-1}(\frac{1}{x})$ for $x > 0$,we have $\cot^{-1}(1 + n(n+1)) = 	an^{-1}(\frac{1}{1 + n(n+1)})$.We can write $\frac{1}{1 + n(n+1)} = \frac{(n+1) - n}{1 + n(n+1)}$.Using the formula $	an^{-1}(x) - 	an^{-1}(y) = 	an^{-1}(\frac{x-y}{1+xy})$,we get $	an^{-1}(n+1) - 	an^{-1}(n)$.Now,the sum is $\sum_{n=1}^{23} (	an^{-1}(n+1) - 	an^{-1}(n))$.This is a telescoping sum: $(	an^{-1}(2) - 	an^{-1}(1)) + (	an^{-1}(3) - 	an^{-1}(2)) + \dots + (	an^{-1}(24) - 	an^{-1}(23))$.This simplifies to $	an^{-1}(24) - 	an^{-1}(1)$.Applying the formula again: $	an^{-1}(\frac{24-1}{1+24	imes 1}) = 	an^{-1}(\frac{23}{25})$.Finally,we need to calculate $\cot(	an^{-1}(\frac{23}{25})) = \cot(\cot^{-1}(\frac{25}{23})) = \frac{25}{23}$.

Column-$I$	Column-$II$
$(a)$ Land snail	$(1)$ Urea
$(b)$ Cartilaginous fishes	$(2)$ Uric acid
$(c)$ Marine fishes	$(3)$ Ammonia

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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