The value of cot left(sumlimits_{n=1}^{50} ta | vedclass

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(A) We know that $	an ^{-1} \left(\frac{x-y}{1+xy}ight) = 	an ^{-1} x - 	an ^{-1} y$. Given the term $	an ^{-1}\left(\frac{1}{1+n+n^{2}}ight)$

Vedclass · Accepted Answer

$\frac{26}{25}$

Vedclass · Answer

(A) We know that $	an ^{-1} \left(\frac{x-y}{1+xy}ight) = 	an ^{-1} x - 	an ^{-1} y$. Given the term $	an ^{-1}\left(\frac{1}{1+n+n^{2}}ight)$,we can rewrite it as: $	an ^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}ight) = 	an ^{-1}(n+1) - 	an ^{-1} n$. Now,consider the summation: $\sum\limits_{n=1}^{50} \left(	an ^{-1}(n+1) - 	an ^{-1} night) = (	an ^{-1} 2 - 	an ^{-1} 1) + (	an ^{-1} 3 - 	an ^{-1} 2) + \dots + (	an ^{-1} 51 - 	an ^{-1} 50)$. This is a telescoping series,so it simplifies to: $	an ^{-1} 51 - 	an ^{-1} 1$. Using the formula $	an ^{-1} x - 	an ^{-1} y = 	an ^{-1} \left(\frac{x-y}{1+xy}ight)$,we get: $	an ^{-1} 51 - 	an ^{-1} 1 = 	an ^{-1} \left(\frac{51-1}{1+51 	imes 1}ight) = 	an ^{-1} \left(\frac{50}{52}ight) = 	an ^{-1} \left(\frac{25}{26}ight)$. Finally,we need to find the value of $\cot \left(	an ^{-1} \left(\frac{25}{26}ight)ight)$. Since $\cot(	an ^{-1} x) = \frac{1}{x}$,we have: $\cot \left(	an ^{-1} \left(\frac{25}{26}ight)ight) = \frac{26}{25}$.

The value of $\cot \left(\sum\limits_{n=1}^{50} \tan ^{-1}\left(\frac{1}{1+n+n^{2}}\right)\right)$ is

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