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MathsJEE Main 6 Apr 2026, Shift 1·Q13·MCQ·Easy–moderate

Inverse Trigonometry · Sum of inverse tangents

Let 0<α<10 < \alpha < 1, β=13α\beta = \dfrac{1}{3\alpha} and tan⁡−1(1−α)+tan⁡−1(1−β)=π4\tan^{-1}(1 - \alpha) + \tan^{-1}(1 - \beta) = \dfrac{\pi}{4}. Then 6(α+β)6(\alpha + \beta) is equal to:
  1. (1)6
  2. (2)7
  3. (3)8
  4. (4)9

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