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Mathematics Question on Inverse Trigonometric Functions

The value of cot⁡(∑n=150tan⁡−1(11+n+n2))\cot\left(\sum_{n=1}^{50} \tan^{-1}\left(\frac{1}{1+n+n^2}\right)\right) is

A

2625\frac{26}{25}

B

2526\frac{25}{26}

C

5051\frac{50}{51}

D

5251\frac{52}{51}

Answer

2625\frac{26}{25}

Explanation

Solution

cot⁡(∑n=150tan⁡−1(11+n+n2))\cot\left(\sum_{n=1}^{50} \tan^{-1}\left(\frac{1}{1+n+n^2}\right)\right)

cot⁡(∑n=150tan⁡−1(n+1−n1+(n+1)n))\cot\left(\sum_{n=1}^{50} \tan^{-1}\left(\frac{n+1-n}{1+(n+1)n}\right)\right)

cot⁡(∑n=150(tan⁡−1(n+1)−tan⁡−1(n)))\cot\left(\sum_{n=1}^{50} (\tan^{-1}(n+1) - \tan^{-1}(n))\right)

cot⁡(tan⁡−1(51)−tan⁡−1(1))\cot(\tan^{-1}(51) - \tan^{-1}(1))

cot⁡(tan⁡−1(51−11+51))\cot(\tan^{-1}(\frac{51-1}{1+51}))

cot⁡(cot⁡−1(5250))\cot(\cot^{-1}(\frac{52}{50}))
=2625=\frac{26}{25}