tan $-$1 a + tan $-$1 b = ${\pi \over 4}$ 0 < a, b < 1 $\Rightarrow {{a + b} \over {1 - ab}} = 1$ a + b = 1 $-$ ab (a + 1)(b + 1) = 2 Now $\left[ {a - {{{a^2}} \over 2} + {{{a^3}} \over 3} + ....} \right] + \left[ {b - {{{b^2}} \over 2} + {{{b^3}} \over 3} + ....} \right]$ $= {\log _e}(1 + a) + {\log _e}(1 + b)$ ($\because$ expansion of log e (1 + x)) $= {\log _e}[(1 + a)(1 + b)]$ $= {\log _e}2$