Let $$a_1 = 1, a_2, a_3, a_4, \ldots$$ be consecutive natural numbers. Then $$\tan^{-1}\left(\frac{1}{1+a_1a_2}\right) + \tan^{-1}\left(\frac{1}{1+a_2a_3}\right) + \ldots + \tan^{-1}\left(\frac{1}{1+a_{2021}a_{2022}}\right)$$ is equal to

The consecutive natural numbers are $$a_1 = 1, a_2 = 2, a_3 = 3, \ldots, a_{2022} = 2022$$.

We use the identity:

$$\tan^{-1}\left(\frac{1}{1 + n(n+1)}\right) = \tan^{-1}(n+1) - \tan^{-1}(n)$$

This follows from the subtraction formula: $$\tan^{-1}(n+1) - \tan^{-1}(n) = \tan^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right) = \tan^{-1}\left(\frac{1}{1+n(n+1)}\right)$$.

So the given sum becomes a telescoping series:

$$\sum_{k=1}^{2021} \tan^{-1}\left(\frac{1}{1+a_k a_{k+1}}\right) = \sum_{k=1}^{2021} \left[\tan^{-1}(k+1) - \tan^{-1}(k)\right]$$

$$= \tan^{-1}(2022) - \tan^{-1}(1) = \tan^{-1}(2022) - \frac{\pi}{4}$$

So, the answer is $$\tan^{-1}(2022) - \frac{\pi}{4}$$.