Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

$\begin{array}{rl}& {\mathrm{T}}_{\mathrm{n}}={\tan }^{-1}\left(\frac{4}{4{\mathrm{n}}^2+3}\right)\\ & {\mathrm{T}}_{\mathrm{n}}={\tan }^{-1}\left(\frac{\left(\mathrm{n}+\frac{1}{2}\right)-\left(\mathrm{n}-\frac{1}{2}\right)}{1+\left(\mathrm{n}+\frac{1}{2}\right)\left(\mathrm{n}-\frac{1}{2}\right)}\right)\\ & {\mathrm{T}}_{\mathrm{n}}={\tan }^{-1}\left(\mathrm{n}+\frac{1}{2}\right)-{\tan }^{-1}\left(\mathrm{n}-\frac{1}{2}\right)\\ & {\mathrm{T}}_1+{\mathrm{T}}_2+\dots +{\mathrm{T}}_{\mathrm{n}}={\tan }^{-1}\left(\mathrm{n}+\frac{1}{2}\right)-{\tan }^{-1}\left(\frac{1}{2}\right)\\ & {\mathrm{S}}_{\infty }=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{1}{2}\right)\end{array}$
option (4)