RM{A}} + {\tan ^{ - 1}}{\rm{B}} = {\tan ^{ - 1}}\LEFT( {\FRAC{{{\rm{A}} + {\rm{B}})}}{{1 - {\rm{AB}}}}} \RIGHT)\) where A > 0, B > 0 & AB < 1 Calculation:Given: \({\tan ^{ - 1}}\left( {1 + {\rm{x}}} \right) + {\tan ^{ - 1}}\left( {1 - {\rm{x}}} \right) = \frac{{\rm{\pi }}}{2}\)We now that, \({\tan ^{ - 1}}{\rm{A}} + {\tan ^{ - 1}}{\rm{B}} = {\tan ^{ - 1}}\left( {\frac{{{\rm{A}} + {\rm{B}})}}{{1 - {\rm{AB}}}}} \right)\) where A > 0, B > 0 & AB < 1\({\tan ^{ - 1}}\left( {1 + {\rm{x}}} \right) + {\tan ^{ - 1}}\left( {1 - {\rm{x}}} \right) = \frac{{\rm{\pi }}}{2}\)\( \Rightarrow {\tan ^{ - 1}}\left( {\frac{{1 + {\rm{x}} + 1 - {\rm{x}}}}{{1 - \left( {1 + {\rm{x}}} \right)\left( {1 - {\rm{x}}} \right)}}} \right) = {\tan ^{ - 1}}\left( {\frac{1}{0}} \right)\)\( \Rightarrow \frac{2}{{1 - 1 + {{\rm{x}}^2}}} = \frac{1}{0}\)⇒ x2 = 0⇒ x = 0