Question

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

• A. \( \frac{26}{25} \)
• B. \( \frac{25}{26} \)
• C. \( \frac{50}{51} \)
• D. \( \frac{52}{51} \)

Hint:

When dealing with sums of inverse tangents, remember to apply the identity for the sum of inverse tangents step by step. This often simplifies the expression significantly.

Answer 1

The Correct Option is A

We are asked to evaluate the following expression: \[ \cot \left( \sum_{n=1}^{50} \tan^{-1} \left( \frac{1}{1 + n + n^2} \right) \right). \] The general identity we will use is the following identity for the sum of inverse tangents: \[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right). \] Using this identity repeatedly, we can simplify the sum. After applying this identity for all the terms in the sum, we obtain the final value of the sum of inverse tangents: \[ \sum_{n=1}^{50} \tan^{-1} \left( \frac{1}{1 + n + n^2} \right) = \tan^{-1} \left( \frac{25}{26} \right). \] Thus, the expression becomes: \[ \cot \left( \tan^{-1} \left( \frac{25}{26} \right) \right). \] Using the identity \( \cot \left( \tan^{-1} x \right) = \frac{1}{x} \), we get: \[ \cot \left( \tan^{-1} \left( \frac{25}{26} \right) \right) = \frac{26}{25}. \] Therefore, the correct answer is \( \frac{26}{25} \).