AP EAMCET · Maths · Inverse Trigonometric Functions
\(\cot \left(\sum_{n=1}^{50} \tan ^{-1}\left(\frac{1}{1+n+n^2}\right)\right)=\)
- 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
\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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