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JEE Mains · Maths · STD 12 - 2. inverse trigonometric function

\(\cot \left(\sum\limits_{n=1}^{50} \tan ^{-1}\left(\frac{1}{1+n+n^{2}}\right)\right)\) का मान है

  1. A \(\frac{26}{25}\)
  2. B \(\frac{25}{26}\)
  3. C \(\frac{50}{51}\)
  4. D \(\frac{52}{51}\)
Verified Solution

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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