The expression $$\tan^{-1}\left(\frac{1}{1 + 1.2}\right) + \tan^{-1}\left(\frac{1}{1 + 2.3}\right) + \tan^{-1}\left(\frac{1}{1 + 3.4}\right) + ........ + \tan^{-1}\left(\frac{1}{1 + n(n + 1)}\right)$$ simplifies to

We will use the formula $$\tan^{-1}\alpha\ -\tan^{-1}\beta\ =\ \tan^{-1}\left(\frac{\alpha\ -\beta\ \ \ }{1+\ \alpha\ \beta\ }\right)$$

The given equation can be re-written as $$\tan^{-1}\left(\ \frac{\ 2-1}{1+1.2}\right)+\tan^{-1}\left(\ \frac{\ 3-2}{1+3.2}\right)+...\ \tan^{-1}\left(\ \frac{\ \left(n+1\right)-n}{1+n\left(n+1\right)}\right)$$

= $$\left(\tan^{-1}2-\tan^{-1}1\right)+\left(\tan^{-1}3-\tan^{-1}2\right)+...\ \left(\tan^{-1}\left(n+1\right)-\tan^{-1}n\right)$$

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