1.
If f(n)=tan⁡[tan⁡−111+2+tan⁡−111+6+tan⁡−111+12+…+tan⁡−111+n(n+1)]f(n) = \tan\left[\tan^{-1} \frac{1}{1+2} + \tan^{-1} \frac{1}{1+6} + \tan^{-1} \frac{1}{1+12} + \ldots + \tan^{-1} \frac{1}{1+n(n+1)}\right] then f(2021)=f(2021) =