KcetInverse Trigonometric FunctionsKCET 2026MCQ+1 / -02026$\tan^{-1}\left(\dfrac{1}{1 + 1 \cdot 2}\right) + \tan^{-1}\left(\dfrac{1}{1 + 2 \cdot 3}\right) + \ldots + \tan^{-1}\left(\dfrac{1}{1 + n(n+1)}\right) = $A$\tan^{-1}\left(\dfrac{n}{n+2}\right)$B$\tan^{-1}\left(\dfrac{n+1}{n}\right)$C$\tan^{-1}\left(\dfrac{n}{n+1}\right)$D$\tan^{-1}\left(\dfrac{n+2}{n}\right)$Check AnswerClear SelectionReveal AnswerShow Explanation