Plz expain how to solve dis...

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ {\tan }^{-1}\left[\frac{\mathrm{x}}{1.2+{\mathrm{x}}^2}\right]+{\tan }^{-1}\left[\frac{\mathrm{x}}{2.3+{\mathrm{x}}^2}\right]+.....+{\tan }^{-1}\left[\frac{\mathrm{x}}{\mathrm{n}\left(\mathrm{n}+1\right)+{\mathrm{x}}^2}\right] \\ \mathrm{By} \mathrm{observing} {\mathrm{n}}^{\mathrm{th}} \mathrm{term}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that} \mathrm{general} \mathrm{term} \mathrm{is} \\ {\mathrm{T}}_{\mathrm{r}}={\tan }^{-1}\left[\frac{\mathrm{x}}{\mathrm{r}\left(\mathrm{r}+1\right)+{\mathrm{x}}^2}\right] \\ \mathrm{Divide} \mathrm{each} \mathrm{term} \mathrm{by} {\mathrm{x}}^2 \\ {\mathrm{T}}_{\mathrm{r}}={\tan }^{-1}\left[\frac{{\displaystyle \frac{\mathrm{x}}{{\mathrm{x}}^2}}}{{\displaystyle \frac{\mathrm{r}\left(\mathrm{r}+1\right)}{{\mathrm{x}}^2}}+{\displaystyle \frac{{\mathrm{x}}^2}{{\mathrm{x}}^2}}}\right] \\ ={\tan }^{-1}\left[\frac{{\displaystyle \frac{1}{\mathrm{x}}}}{{\displaystyle 1+\frac{\mathrm{r}\left(\mathrm{r}+1\right)}{{\mathrm{x}}^2}}}\right] \\ ={\tan }^{-1}\left[\frac{{\displaystyle \frac{\mathrm{r}+1}{\mathrm{x}}-\frac{\mathrm{r}}{\mathrm{x}}}}{{\displaystyle 1+\frac{\mathrm{r}+1}{\mathrm{x}}.\frac{\mathrm{r}}{\mathrm{x}}}}\right] \\ {\mathrm{T}}_{\mathrm{r}}={\tan }^{-1}\left(\frac{\mathrm{r}+1}{\mathrm{x}}\right)-{\tan }^{-1}\left(\frac{\mathrm{r}}{\mathrm{x}}\right) \left[{\tan }^{-1}\left(\frac{\mathrm{A}-\mathrm{B}}{1+\mathrm{AB}}\right)={\tan }^{-1}\mathrm{A}-{\tan }^{-1}\mathrm{B}\right] \\ {\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{r}=1}^{\mathrm{n}} \left\{{\tan }^{-1}\left(\frac{\mathrm{r}+1}{\mathrm{x}}\right)-{\tan }^{-1}\left(\frac{\mathrm{r} }{\mathrm{x}}\right)\right\} \\ ={\tan }^{-1}\left(\frac{2}{\mathrm{x}}\right)-{\tan }^{-1}\left(\frac{1}{\mathrm{x}}\right)+{\tan }^{-1}\left(\frac{3}{\mathrm{x}}\right)-{\tan }^{-1}\left(\frac{2}{\mathrm{x}}\right)+{\tan }^{-1}\left(\frac{4}{\mathrm{x}}\right)-{\tan }^{-1}\left(\frac{3}{\mathrm{x}}\right)+...+{\tan }^{-1}\left(\frac{\mathrm{n}+1}{\mathrm{x}}\right)-{\tan }^{-1}\left(\frac{\mathrm{n}}{\mathrm{x}}\right) \\ =\cancel{{\tan }^{-1}\left(\frac{2}{\mathrm{x}}\right)}-{\tan }^{-1}\left(\frac{1}{\mathrm{x}}\right)+\cancel{{\tan }^{-1}\left(\frac{3}{\mathrm{x}}\right)}-\cancel{{\tan }^{-1}\left(\frac{2}{\mathrm{x}}\right)}+\cancel{{\tan }^{-1}\left(\frac{4}{\mathrm{x}}\right)}-\cancel{{\tan }^{-1}\left(\frac{3}{\mathrm{x}}\right)}+...+{\tan }^{-1}\left(\frac{\mathrm{n}+1}{\mathrm{x}}\right)-\cancel{{\tan }^{-1}\left(\frac{\mathrm{n}}{\mathrm{x}}\right)} \\ ={\tan }^{-1}\left(\frac{\mathrm{n}+1}{\mathrm{x}}\right)-{\tan }^{-1}\left(\frac{1}{\mathrm{x}}\right) \\ ={\cot }^{-1}\left(\frac{\mathrm{x}}{\mathrm{n}+1}\right)-{\cot }^{-1}\left(\mathrm{x}\right) \left[\mathrm{As} {\tan }^{-1}\mathrm{A}={\cot }^{-1}\frac{1}{\mathrm{A}}\right] \\ =\frac{\mathrm{\pi }}{2}-{\tan }^{-1}\left(\frac{\mathrm{x}}{\mathrm{n}+1}\right)-\left[\frac{\mathrm{\pi }}{2}-{\tan }^{-1}\left(\mathrm{x}\right)\right] \left[\mathrm{As} {\cot }^{-1}\mathrm{A}=\frac{\mathrm{\pi }}{2}-{\tan }^{-1}\mathrm{A}\right] \\ \mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\mathbf{x}\right)\mathbf{-}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\frac{\mathbf{x}}{\mathbf{n}\mathbf{+}\mathbf{1}}\right) \end{gathered}$$

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