JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
The value of \(\cot \left(\sum\limits_{n=1}^{50} \tan ^{-1}\left(\frac{1}{1+n+n^{2}}\right)\right)\) is
- A \(\frac{26}{25}\)
- B \(\frac{25}{26}\)
- C \(\frac{50}{51}\)
- D \(\frac{52}{51}\)
Answer & Solution
Correct Answer
(A) \(\frac{26}{25}\)
Step-by-step Solution
Detailed explanation
\(\tan ^{-1} \frac{1}{1+n+n^{2}}=\tan ^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right)\) \(=\tan ^{-1}(n+1)-\tan ^{-1} n\) so, \(\sum\limits_{n=1}^{50}\left(\tan ^{-1}(n+1)-\tan ^{-1} n\right)\) \(=\tan ^{-1} 51-\tan ^{-1} 1\)…
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