MCQ
The value of $\cot \left(\sum\limits_{n=1}^{50} \tan ^{-1}\left(\frac{1}{1+n+n^{2}}\right)\right)$ is
- ✓$\frac{26}{25}$
- B$\frac{25}{26}$
- C$\frac{50}{51}$
- D$\frac{52}{51}$
✓
Answer
Correct option: A.
$\frac{26}{25}$
a
$\tan ^{-1} \frac{1}{1+n+n^{2}}=\tan ^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right)$
$\tan ^{-1} \frac{1}{1+n+n^{2}}=\tan ^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right)$
$=\tan ^{-1}(n+1)-\tan ^{-1} n$
so, $\sum\limits_{n=1}^{50}\left(\tan ^{-1}(n+1)-\tan ^{-1} n\right)$
$=\tan ^{-1} 51-\tan ^{-1} 1$
$\cot \left(\sum\limits_{n=1}^{50} \tan ^{-1}\left(\frac{1}{1+n+n^{2}}\right)\right)=\cot \left(\tan ^{-1} 51+\tan ^{-1} 1\right)$
$=\frac{1}{\tan \left(\tan ^{-1} 51-\tan ^{-1} 1\right)}=\frac{1+51 \times 1}{51-1}=\frac{52}{50}=\frac{26}{25}$
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