Verified Solution • 2025–26 Session moderate

QJEEM16Ma506b: JEE Main 2016 Mathematics - Class 11 JEEM16Ma506b Solution: If   A > 0, B > 0   and    A + B…

If   A > 0, B > 0   and    A + B = π6{\pi \over 6} ,

then the minimum value of tanA + tanB is :

Multiple Choice Options

  • A 3−2\sqrt 3 - \sqrt 2
  • B 2−32 - \sqrt 3
  • C 4−234 - 2\sqrt 3
  • D 23{2 \over {\sqrt 3 }}

Step-by-Step Verified Solution

Given,

A + B = π6{\pi \over 6}

∴\therefore    tan(A + B) = tan (π6)\left( {{\pi \over 6}} \right) = 13{1 \over {\sqrt 3 }}

We know,

tan(A + B) = tan⁡A+tan⁡B1−tan⁡Atan⁡B{{\tan A + \tan B} \over {1 - \tan A\tan B}}

⇒\Rightarrow    13{1 \over {\sqrt 3 }} = y1−tan⁡Atan⁡B{y \over {1 - \tan A\tan B}}

where y = tan A + tan B

⇒\Rightarrow     tanA tanB = 1 −- 3\sqrt 3 y

Also AM    ≥\ge   GM

⇒\Rightarrow     tan⁡A+tan⁡B2≥tan⁡Atan⁡B{{\tan A + \tan B} \over 2} \ge \sqrt {\tan A\tan B}

⇒\Rightarrow    y  ≥\ge  2 1−3y\sqrt {1 - \sqrt 3 y}

⇒\Rightarrow    y2 ≥\ge 4 −- 4 3y{\sqrt 3 y}

⇒\Rightarrow     y2 + 4 3y{\sqrt 3 y} −- 4 ≥\ge 0

⇒\Rightarrow    y   ≤\le   −- 2 3\sqrt 3 −- 4   

or   y ≥\ge −- 2 3\sqrt 3 + 4

(y ≤\le −- 2 3\sqrt 3 −- 4 is not possible as tan B > 0)

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