If tan - 1 1 1 + 1 . 2 + tan - 1 1 1 + 2 . 3 + ⋯ + tan - 1 1 1 + n ( n + 1 ) = tan - 1 [ x ] , then x =

We have, ${\tan }^{-1}\left[\frac{1}{1+1.2}\right]={\tan }^{-1}\left[\frac{2-1}{1+1.2}\right]$

Now, using the formula ${\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{x-y}{1+xy}\right), xy>-1$

We get, ${\tan }^{-1}\left[\frac{1}{1+1.2}\right]={\tan }^{-1}2-{\tan }^{-1}1$

Similarly, ${\tan }^{-1}\left[\frac{1}{1+2.3}\right]={\tan }^{-1}3-{\tan }^{-1}2$

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

$\Rightarrow {\tan }^{-1}\left[\frac{1}{1+1.2}\right]+{\tan }^{-1}\left[\frac{1}{1+2.3}\right]+\cdots +{\tan }^{-1}\left[\frac{1}{1+n(n+1)}\right]={\tan }^{-1}2-{\tan }^{-1}1+{\tan }^{-1}3-{\tan }^{-1}2+\cdots +{\tan }^{-1}\left(n+1\right)-{\tan }^{-1}n$

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

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

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

$\Rightarrow {\tan }^{-1}\left[\frac{n}{2+n}\right]={\tan }^{-1}\left[x\right]$

$\Rightarrow x=\frac{n}{2+n}$