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  • The value of    <img alt="\mathrm{\cot \left(\sum_{n=1}^{50} \tan ^{-1}\left(\frac{1}{1+n+n^{2}}\right)\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Ccot%20

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

Option: 1

\frac{26}{25}


Option: 2

\frac{25}{26}


Option: 3

\frac{50}{51}


Option: 4

\frac{52}{51}


Answers (1)

best_answer

\mathrm{\sum_{n=1}^{50} \tan ^{-1}\left(\frac{(n+1)-(n)}{1+(n+1)(n)}\right)=\sum_{n=1}^{50}\left(\tan ^{-1}(n+1)-\tan ^{-1}(n)\right] }\\

\mathrm{\left.=\left(\tan ^{-1} 2-\tan ^{-1}1\right)+\left(\tan ^{-1} 3-\tan ^{-1} 2\right)+\ldots-(\tan ^{-1} 51-\tan ^{-1} 50\right)} \\

\mathrm{=\tan ^{-1}(51)-\tan ^{-1}(1)} \\

\mathrm{=\tan ^{-1} \frac{51-1}{1+51 \times 1}=\tan ^{-1} \frac{50}{52}=\tan ^{-1} \frac{25}{26}} \\

\mathrm{\text { so } \cot \left(\tan ^{-1}\left(\frac{25}{26}\right)\right)=\frac{26}{25} }

Hence the correct answer is option 1

Posted by

HARSH KANKARIA

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