cot (∑_n=1^50 tan^-1 (1/(1+n+n^2) ) )=

Correct answer: (A) 26/25. cot (∑_n=1^50 tan^-1 (1/(1+n+n^2) ) ) = cot (∑_n=1^50 tan^-1 ((n+1-n)/(1+(n+1) n) ) ) = cot (∑_n=1^50 (tan^-1(n+1)-tan^-1 n ) ) = cot (tan^-1 51-tan^-1 1 )=cot (tan^-1 (51-1)/(1+51) ) = cot (cot^-1 52/50 )=52/50=26/25

Q64 / 160 · AP EAMCET 2024 (22 May Shift 2) · Mathematics

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

Correct answer
A
$\frac{26}{25}$
B
$\frac{25}{26}$
C
$\frac{50}{51}$
D
$\frac{52}{51}$

Solution

$\begin{aligned} & \cot \left(\sum_{n=1}^{50} \tan ^{-1}\left(\frac{1}{1+n+n^2}\right)\right) \\ = & \cot \left(\sum_{n=1}^{50} \tan ^{-1}\left(\frac{n+1-n}{1+(n+1) n}\right)\right) \\ = & \cot \left(\sum_{n=1}^{50}\left(\tan ^{-1}(n+1)-\tan ^{-1} n\right)\right) \\ = & \cot \left(\tan ^{-1} 51-\tan ^{-1} 1\right)=\cot \left(\tan ^{-1} \frac{51-1}{1+51}\right) \\ = & \cot \left(\cot ^{-1} \frac{52}{50}\right)=\frac{52}{50}=\frac{26}{25}\end{aligned}$

Hence, the correct answer is option: 'A'.

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

Solution

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