If 0<a, b<1, and tan-1a+tan-1b=π4, then the value a+b-a2+b22+a3+b33-a4+b44+···is

tan-1a+b1-ab=π4 ⇒ a+b1-ab=1 a+b+ab=1 ⇒ (1+a) (1+b)=2 a-a22+a33-a44....+b-b22+b33.... log(1+x)=x-x22+x33-x44---- log(1+a)+log(1+b)=log(1+a) (1+b) =log 2

If 0<a, b<1, and tan-1a+tan-1b=π4, then the value a+b-a2+b22+a3+b33-a4+b44+···is

tan-1a+b1-ab=π4 ⇒ a+b1-ab=1 a+b+ab=1 ⇒ (1+a) (1+b)=2 a-a22+a33-a44....+b-b22+b33.... log(1+x)=x-x22+x33-x44---- log(1+a)+log(1+b)=log(1+a) (1+b) =log 2

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If $0 < a, b < 1,$ and $\tan^{- 1} a + \tan^{- 1} b = \frac{π}{4},$ then the value $(a + b) - (\frac{a^{2} + b^{2}}{2}) + (\frac{a^{3} + b^{3}}{3}) - (\frac{a^{4} + b^{4}}{4}) + \cdot \cdot \cdot$ is

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${lof}_{e} (\frac{e}{2})$

$\log_{e} 2$

$e^{2} - 1$

answer is C.

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Detailed Solution

$\tan^{- 1} (\frac{a + b}{1 - ab}) = \frac{π}{4} \Rightarrow \frac{a + b}{1 - ab} = 1 a + b + ab = 1 \Rightarrow (1 + a) (1 + b) = 2 (a - \frac{a^{2}}{2} + \frac{a^{3}}{3} - \frac{a^{4}}{4} . . . .) + (b - \frac{b^{2}}{2} + \frac{b^{3}}{3} . . . .) \log (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} - - - - \log (1 + a) + \log (1 + b) = \log (1 + a) (1 + b) = \log 2$