If ^ -1 [(1)/(1+1.2)]+ ^ -1 [(1)/(1+2.3)]+ + ^ -1 [(1)/(1+n(n+1))]=...

$$\begin{aligned} & \text{ Given } \tan ^{-1}\left(\frac{1}{1+1.2}\right)+\tan ^{-1}\left(\frac{1}{1+2.3}\right)+\ldots \\ & +\tan ^{-1}\left(\frac{1}{1+n(n+1)}\right)=\tan ^{-1}(x)\end{aligned}$$$$\begin{aligned} \Rightarrow \tan ^{-1}\left(\frac{2-1}{1+1.2}\right) & +\tan ^{-1}\left(\frac{3-2}{1+2.3}\right)+\ldots \\ & +\tan ^{-1}\left(\frac{n+1-n}{1+n(n+1)}\right)=\tan ^{-1}(x)\end{aligned}$$$$\begin{aligned} & \Rightarrow \tan ^{-1}(2)-\tan ^{-1}(1)+\tan ^{-1}(3)-\tan ^{-1}(2)+\ldots \\ & +\tan ^{-1}(\mathrm{n}+1)-\tan ^{-1}(\mathrm{n})=\tan ^{-1}(x)\end{aligned}$$\Rightarrow \tan ^{-1}(\mathrm{n}+1)-\tan ^{-1}(1)=\tan ^{-1}(x)\left\{\because \tan ^{-1}(A)-\tan ^{-1}(B)=\tan ^{-1}\left(\frac{A-B}{1+A \cdot B}\right)\right\}\Rightarrow \tan ^{-1}\left(\frac{n}{n+2}\right)=\tan ^{-1}(x) \Rightarrow x=\frac{n}{n+2}$