Given,
tan−1(1+a1a21)+tan−1(1+a2a31)................tan−1(1+a2021a20221)
=tan−1(1+a1a2a2−a1)+tan−1(1+a2a3a3−a2)................tan−1(1+a2021a2022a2022−a2021)
=tan−1a2−tan−1a1+tan−1a3−tan−1a2..........+tan−1a2022−tan−1a2021
using the formula tan−1x−tan−1y=tan−1(1+xyx−y)
=tan−1a2022−tan−1a1
=tan−1a2022−tan−11as givena1=1
=tan−1a2022−4π
=2π−cot−1a2022−4π
=4π−cot−1a2022
=4π−cot−12022
As a1=1, a2=2 .... and so on a2022=2022