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If $\tan^{- 1} A + \tan^{- 1} B = \frac{π}{4},$ where $0 < A, B < 1,$ then the value of
$(A + B) - (\frac{A^{2} + B^{2}}{2}) + (\frac{A^{3} + B^{3}}{3}) - (\frac{A^{4} + B^{4}}{4}) + \dots \infty,$ is

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detailed solution

Correct option is C

We have,tan−1⁡A+tan−1⁡B=π4⇒tan−1⁡A+B1−AB=π4⇒A+B1−AB=tan⁡π4⇒A+B1−AB=1⇒A+B=1−AB⇒AB+A+B=1⇒1+A+B+AB=2⇒(1+A)(1+B)=2∴(A+B)−A2+B22+A3+B33−A4+B44+…=A−12A2+13A3−14A4+…+B−12B2+13B3−14B4+…=lne⁡(1+A)+lne⁡(1+B)=lne⁡(1+A)(1+B)=lne⁡2=ln⁡2

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