If tan⁻¹(1/(1+1· 2)) + tan⁻¹(1/7) + … + tan⁻¹(1/(111)) = S, then find tan S.

$\tan^{-1}\left(\frac{1}{1+1\cdot2}\right) + \tan^{-1}\left(\frac{1}{1+2\cdot3}\right) + \dots + \tan^{-1}\left(\frac{1}{1+10\cdot11}\right) = S$ $\therefore \tan^{-1}\left(\frac{2-1}{1+2\cdot1}\right) + \tan^{-1}\left(\frac{3-2}{1+3\cdot2}\right) + \dots + \tan^{-1}\left(\frac{11-10}{1+11\cdot10}\right) = S$ $(\tan^{-1}2 - \tan^{-1}1) + (\tan^{-1}3 - \tan^{-1}2) + \dots + (\tan^{-1}11 - \tan^{-1}10) = S$ $\tan^{-1}11 - \tan^{-1}1 = S$ $\tan^{-1}\frac{10}{12} = S$ $\frac{5}{6} = \tan S$