JEE MAIN - Mathematics Hindi (2022 - 27th June Evening Shift - No. 13)

$$\cot \left(\sum\limits_{n=1}^{50} \tan ^{-1}\left(\frac{1}{1+n+n^{2}}\right)\right)$$ का मान है-

$$\frac{26}{25}$$

$$\frac{25}{26}$$

$$\frac{50}{51}$$

$$\frac{52}{51}$$