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AP EAMCET · Maths · Inverse Trigonometric Functions

\(\cot \left(\sum_{n=1}^{50} \tan ^{-1}\left(\frac{1}{1+n+n^2}\right)\right)=\)

  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

\begin{aligned} & \cot \left(\sum_{n=1}^{50} \tan ^{-1}\left(\frac{1}{1+n+n^2}\right)\right) \\ = & \cot \left(\sum_{n=1}^{50} \tan ^{-1}\left(\frac{n+1-n}{1+(n+1) n}\right)\right) \\ = & \cot \left(\sum_{n=1}^{50}\left(\tan ^{-1}(n+1)-\tan ^{-1} n\right)\right) \\ = & \cot…

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