Inverse Trigonometric FunctionsmediumPYQ · JEE Main · 6 Apr 2026 · Shift 1 (Morning)Free

Inverse-Tangent Equation: 6(α+β) = 7 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
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
A66
B77correct
C88
D99
Solution
Step 1: tan⁡(tan⁡−1p+tan⁡−1q)=p+q1−pq\tan\big(\tan^{-1}p+\tan^{-1}q\big)=\dfrac{p+q}{1-pq} with p=1−α, q=1−βp=1-\alpha,\ q=1-\beta, tan⁡π4=1\tan\dfrac{\pi}{4}=1:
(1−α)+(1−β)1−(1−α)(1−β)=1.\dfrac{(1-\alpha)+(1-\beta)}{1-(1-\alpha)(1-\beta)}=1.
 Step 2: (1−α)+(1−β)=1−(1−α)(1−β)(1-\alpha)+(1-\beta)=1-(1-\alpha)(1-\beta). LHS =2−α−β=2-\alpha-\beta; (1−α)(1−β)=1−α−β+αβ⇒(1-\alpha)(1-\beta)=1-\alpha-\beta+\alpha\beta\Rightarrow RHS =α+β−αβ=\alpha+\beta-\alpha\beta.
⇒2−α−β=α+β−αβ.\Rightarrow2-\alpha-\beta=\alpha+\beta-\alpha\beta.
 Step 3: 2+αβ=2(α+β)2+\alpha\beta=2(\alpha+\beta). Step 4: β=13α⇒αβ=13\beta=\dfrac{1}{3\alpha}\Rightarrow\alpha\beta=\dfrac13.
2(α+β)=2+13=73⇒α+β=76.2(\alpha+\beta)=2+\dfrac13=\dfrac73\Rightarrow\alpha+\beta=\dfrac{7}{6}.
 Step 5: ∴6(α+β)=6⋅76=7\therefore6(\alpha+\beta)=6\cdot\dfrac76=7. Correct answer: (2)
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