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.