Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

$S={\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{13}+\dots \dots$ up to$10$ terms.

$\mathrm{S}={\tan }^{-1}\left(\frac{2-1}{1+1.2}\right)+{\tan }^{-1}\left(\frac{3-2}{1+2.3}\right)+{\tan }^{-1}\left(\frac{4-3}{1+3.4}\right)+\dots \dots +{\tan }^{-1}\left(\frac{11-10}{1+11.10}\right)$

$\Rightarrow \mathrm{S}=\left({\tan }^{-1}2-{\tan }^{-1}1\right)+\left({\tan }^{-1}3-{\tan }^{-1}2\right)+\dots \dots +\left({\tan }^{-1}11-{\tan }^{-1}10\right)$

$\Rightarrow S={\tan }^{-1}11-{\tan }^{-1}1$

$\Rightarrow S={\tan }^{-1}\left(\frac{11-1}{1+11·1}\right)={\tan }^{-1}\left(\frac{10}{12}\right)$

$\therefore \tan \left(S\right)=\frac{5}{6}$