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tan^-1 a + tan^-1 b = pi/4 gives (1 + a)(1 + b) = 2

This type of question, asked 2 times: 2021 Paper 26, 6 Apr 2026 S1. All 2 questions below, oldest first.

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  1. Question 1

    2021 Paper 26

    68
    If 0<a,b<10 < a,b < 1, and tan⁡−1a+tan⁡−1b=π4\tan^{-1}a +\tan^{-1}b =\frac{\pi}{4}, then the value of (a+b)−(a2+b22)+(a3+b33)−(a4+b44)+…(a + b) -\left( \frac{a^{2}+b^{2}}{2} \right)+\left( \frac{a^{3}+b^{3}}{3} \right)-\left( \frac{a^{4}+b^{4}}{4} \right)+ \ldots is:

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  2. Question 2

    6 Apr 2026 S1

    63
    Let 0<α<1,β=13α0 <\alpha< 1,\beta=\frac{1}{3\alpha} and tan⁡−1(1−α)+tan⁡−1(1−β)=π4\tan^{- 1}(1 -\alpha) +\tan^{- 1}(1 -\beta) =\frac{\pi}{4}. Then 6(α+β)6(\alpha+\beta) is equal to:

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