Let a₁ = 1, a₂, a₃, a₄, ........ be consecutive natural numbers.
Then ${\tan }^{-1}\left(\frac{1}{1+{\mathrm{a}}_1{\mathrm{a}}_2}\right)+{\tan }^{-1}\left(\frac{1}{1+{\mathrm{a}}_2{\mathrm{a}}_3}\right)+\dots ..+{\tan }^{-1}\left(\frac{1}{1+{\mathrm{a}}_{2021}{\mathrm{a}}_{2022}}\right)$ is equal to

$$\begin{gathered} \begin{array}{l}{\tan }^{-1}\left(\frac{{\mathrm{a}}_2-{\mathrm{a}}_1}{1+{\mathrm{a}}_1{\mathrm{a}}_2}\right)+\dots ..+{\tan }^{-1}\left(\frac{{\mathrm{a}}_{2022}-{\mathrm{a}}_{2021}}{1+{\mathrm{a}}_{2021}{\mathrm{a}}_{2022}}\right)\\ ={\tan }^{-1}{\mathrm{a}}_2-{\tan }^{-1}{\mathrm{a}}_1+\dots ..+{\tan }^{-1}\left({\mathrm{a}}_{2022}\right)-{\tan }^{-1}\left({\mathrm{a}}_{2021}\right)\\ ={\tan }^{-1}\left({\mathrm{a}}_{2022}\right)-{\tan }^{-1}{\mathrm{a}}_1={\tan }^{-1}\left({\mathrm{a}}_{2022}\right)-\frac{\mathrm{\pi }}{4} \\ =\frac{\mathrm{\pi }}{2}-{\cot }^{-1}\left({\mathrm{a}}_{2022}\right)-\frac{\mathrm{\pi }}{4}\left(\because {\mathrm{a}}_1=1\right) \\ =\frac{\mathrm{\pi }}{4}-{\cot }^{-1}\left({\mathrm{a}}_{2022}\right)=\frac{\mathrm{\pi }}{4}-{\cot }^{-1}\left(2022\right) \\ Ans: A,C\end{array} \end{gathered}$$