cot left(sum_{n=1}^{50} tan ^{-1}left(frac{1} | vedclass

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

Vedclass · Accepted Answer

$\frac{26}{25}$

Vedclass · Answer

(A) We know that $	an^{-1} x - 	an^{-1} y = 	an^{-1} \left( \frac{x-y}{1+xy} ight)$.We can rewrite the general term as $	an^{-1} \left( \frac{(n+1)-n}{1+n(n+1)} ight) = 	an^{-1}(n+1) - 	an^{-1} n$.Thus,the sum is $\sum_{n=1}^{50} (	an^{-1}(n+1) - 	an^{-1} n)$.This is a telescoping sum: $(	an^{-1} 2 - 	an^{-1} 1) + (	an^{-1} 3 - 	an^{-1} 2) + \dots + (	an^{-1} 51 - 	an^{-1} 50) = 	an^{-1} 51 - 	an^{-1} 1$.Using the formula $	an^{-1} x - 	an^{-1} y = 	an^{-1} \left( \frac{x-y}{1+xy} ight)$,we get $	an^{-1} \left( \frac{51-1}{1+51 	imes 1} ight) = 	an^{-1} \left( \frac{50}{52} ight) = 	an^{-1} \left( \frac{25}{26} ight)$.Finally,$\cot \left( 	an^{-1} \left( \frac{25}{26} ight) ight) = \cot \left( \cot^{-1} \left( \frac{26}{25} ight) ight) = \frac{26}{25}$.

$\cot \left(\sum_{n=1}^{50} \tan ^{-1}\left(\frac{1}{1+n+n^2}\right)\right) = $

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