Let $a_1=1,a_2,a_3,a_4,\mathrm{.........}$ be consecutive natural numbers then $Ta{n}^{-1}\left(\frac{1}{1+a_1{a}_2}\right)+Ta{n}^{-1}\left(\frac{1}{1+a_2{a}_3}\right)+\mathrm{.......}+Ta{n}^{-1}\left(\frac{1}{1+a_{2021}{a}_{2022}}\right)$ is equal to 

$Ta{n}^{-1}\left(\frac{a_2-a_1}{1+a_1{a}_2}\right)+\mathrm{...........}$

$Ta{n}^{-1}\left(\frac{a_{2022}-a_{2021}}{1+a_{2021}{a}_{2022}}\right)$

$Ta{n}^{-}\left(a_2\right)-Ta{n}^{-1}{a}_1+\mathrm{.....}$

$Ta{n}^{-1}{a}_{2022}-Ta{n}^{-1}\left(a_{2021}\right)$

$=Ta{n}^{-1}{a}_{2022}-Ta{n}^{-1}1$

$=\pi /2-{\cot }^{-1}{a}_{2022}-\frac{\pi }{4}$

$=\frac{\pi }{4}-{\cot }^{-1}{a}_{2022}$