JEE Main 2020 (Online) 5th September Morning ShiftQuestion 1
If $S$ is the sum of the first $10$ terms of the series $\tan^{-1}\left(\frac13\right)+\tan^{-1}\left(\frac17\right)+\tan^{-1}\left(\frac1{13}\right)+\tan^{-1}\left(\frac1{21}\right)+\ldots$, then $\tan(S)$ is equal to :
Inverse Trigonometric Functions
Solution
$\frac1{n^2+n+1}=\frac{(n+1)-n}{1+n(n+1)}\Rightarrow$ term $=\tan^{-1}(n+1)-\tan^{-1}n$. $S=\tan^{-1}11-\tan^{-1}1$. $\tan S=\frac{11-1}{1+11}=\frac{10}{12}=\frac56$. Option A.