If ${\tan }^{-1} \frac{1}{1+2}+{\tan }^{-1} \frac{1}{1+\left(2\right)\left(3\right)}+$${\tan }^{-1} \frac{1}{1+\left(3\right)\left(4\right)}+\dots +{\tan }^{-1} \frac{1}{1+n(n+1)}={\tan }^{-1} \theta$

then $\theta$=

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

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

so that L.H.S. of the given equation is 

$\begin{array}{l}{\tan }^{-1} 2-{\tan }^{-1} 1+{\tan }^{-1} 3-{\tan }^{-1} 2+\cdots +\\ {\tan }^{-1} (n+1)-{\tan }^{-1} n\\ ={\tan }^{-1} (n+1)-{\tan }^{-1} 1={\tan }^{-1} \frac{n+1-1}{1+(n+1)}\\ ={\tan }^{-1} \frac{n}{n+2}\end{array}$

so that ${\tan }^{-1} \frac{n}{n+2}={\tan }^{-1} \theta \Rightarrow \theta =\frac{n}{n+2}$