Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

We know that, $$\begin{array}{r}{\tan }^{-1}\frac{1}{1+2}+{\tan }^{-1}\frac{1}{1+2\times 3}+{\tan }^{-1}\frac{1}{1+3\times 4}+\dots \dots +\\ {\tan }^{-1}\frac{1}{1+(n-1)n}+{\tan }^{-1}\frac{1}{1+n(n+1)}+\dots \dots +\\ {\tan }^{-1}\frac{1}{1+(n+19)(n+20)}={\tan }^{-1}\frac{n+19}{n+21}\\ \Rightarrow {\tan }^{-1}\frac{n-1}{n+1}+{\tan }^{-1}\frac{1}{1+n(n+1)}+{\tan }^{-1}\frac{1}{1+(n+1)(n+2)}\\ +\dots \dots +\frac{1}{1+(n+19)(n+20)}={\tan }^{-1}\frac{n+19}{n+21}\\ {\tan }^{-1}\frac{1}{1+n(n+1)}+{\tan }^{-1}\frac{1}{1+(n+1)(n+2)}+\dots \dots +\\ \frac{1}{1+(n+19)(n+20)}={\tan }^{-1}\frac{n+19}{n+21}-{\tan }^{-1}\frac{n-1}{n+1}\end{array}$$ $$\begin{array}{rl}& {\tan }^{-1}\left(\frac{1}{n^2+n+1}\right)+{\tan }^{-1}\left(\frac{1}{n^2+3n+3}\right)+\dots \dots +\\ & ={\tan }^{-1}\left(\frac{\frac{n+19}{n+21}-\frac{n-1}{n+1}}{1+\frac{n+19}{n+21}\times \frac{n-1}{n+1}}\right)\\ & ={\tan }^{-1}\frac{1}{1+(n+19)(n+20)}\frac{20}{n^2+20n+1}=\mathrm{S}\\ & \therefore {\tan }^{-1}\mathrm{S}=\frac{20}{n^2+20n+1}\end{array}$$