$a_{1}=1, a_{2}, a_{3}, a_{4}, \ldots .$. be consecutive natural numbers. $\begin{aligned} & \therefore \quad a_2=2, a_3=3, \ldots ., a_{2021}=2021, a_{2022}=2022 \\\\ & \tan ^{-1}\left[\frac{1}{1+a_1 a_2}\right]=\tan ^{-1}\left[\frac{1}{1+1 \cdot 2}\right]=\tan ^{-1}(1)-\tan ^{-1}\left(\frac{1}{2}\right) \\\\ & \tan ^{-1}\left[\frac{1}{1+a_2 a_3}\right]=\tan ^{-1}\left[\frac{1}{1+2 \cdot 3}\right]=\tan ^{-1}\left(\frac{1}{2}\right)-\tan ^{-1}\left(\frac{1}{3}\right) \end{aligned}$ . . . . $\begin{aligned} \tan ^{-1}\left[\frac{1}{1+a_{2021} a_{2022}}\right] & =\tan ^{-1}\left[\frac{1}{1+2021 \cdot 2022}\right] \\\\ & =\tan ^{-1}\left(\frac{1}{2021}\right)-\tan ^{-1}\left(\frac{1}{2022}\right) \end{aligned}$ $\therefore$ $\tan ^{-1}\left(\frac{1}{1+a_{1} a_{2}}\right)+\tan ^{-1}\left(\frac{1}{1+a_{2} a_{3}}\right)+\ldots . .+\tan ^{-1}\left(\frac{1}{1+a_{2021} a_{2022}}\right)$ $=\tan ^{-1}(1)-\tan ^{-1}\left(\frac{1}{2022}\right)=\frac{\pi}{4}-\cot ^{-1}(2022)$ $=\frac{\pi}{4}-\left(\frac{\pi}{2}-\tan ^{-1}(2022)\right)=\tan ^{-1}(2022)-\frac{\pi}{4}$