Question

If \( \tan^{-1}\left(\frac{1}{1+1\cdot2}\right) + \tan^{-1}\left(\frac{1}{1+2\cdot3}\right) + \ldots + \tan^{-1}\left(\frac{1}{1+n(n+1)}\right) = \tan^{-1}(x) \), then \( x \) is equal to:

• A. \(\frac{1}{n+1}\)
• B. \(\frac{n}{n+1}\)
• C. \(\frac{1}{n+2}\)
• D. \(\frac{n}{n+2}\)

Hint:

Telescoping series in trigonometric identities often simplify to terms involving only the first and last elements.

Answer 1

The Correct Option is D

Given the series: \[ \tan^{-1} \left( \frac{1}{1 + 1 \cdot 2} \right) + \tan^{-1} \left( \frac{1}{1 + n(n+1)} \right) = \tan^{-1}(x) \] We can simplify the terms as follows: \[ \tan^{-1} \left( \frac{2 - 1}{1 + 1 \cdot 2} \right) + \tan^{-1} \left( \frac{3 - 2}{1 + 2 \cdot 3} \right) + \cdots \] Using the difference identity for inverse tangents: \[ \tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left( \frac{A - B}{1 + AB} \right) \] After applying this identity for each pair, we simplify the entire expression to: \[ \tan^{-1}(n+1) - \tan^{-1}(1) = \tan^{-1}(x) \] Thus, we find: \[ \tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left( \frac{A - B}{1 + A \cdot B} \right) \] \[ \Rightarrow \tan^{-1} \left( \frac{n}{n+2} \right) = \tan^{-1}(x) \] Finally, we conclude that: \[ x = \frac{n}{n + 2} \]