Mht Cet
Inverse Trigonometric Functions
MHT CET 2026 18th April Morning Shift
MCQ+2 / -02026
If $\sum\limits_{n=1}^{2026}\tan^{-1}\left(\dfrac{1}{n^2+n+1}\right) = \tan^{-1}\left(1 - \dfrac{1}{x}\right)$, where $x \neq 0$, then $x = $
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