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Question

If 0 < a, b < 1, and tan-1a + tan-1b = $${\pi  \over 4}$$, then the value of

$$(a + b) - \left( {{{{a^2} + {b^2}} \over 2}} \right) + \left( {{{{a^3} + {b^3}} \over 3}} \right) - \left( {{{{a^4} + {b^4}} \over 4}} \right) + .....$$ is :

Accepted Answer

tan-1a + tan-1b = $${\pi  \over 4}$$ 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 loge(1 + x))

$$ = {\log _e}[(1 + a)(1 + b)]$$

$$ = {\log _e}2$$

Solutions · 1