If A>0,B>0,A+B=π/3 and maximum value of tan A tan B is M then the value of 1/M is
A+B=π/3⇒ tan(A+B)=3⇒ tanA+tanB1−tanAtanB=3⇒ tanA+ytanA1−y=3 (where y=tanAtanB)⇒ tan2A+3(y−1)tanA+y=0
For real values of tan A,
3(y−1)2−4y≥0⇒3y2−10y+3≥0⇒(y−3)(3y−1)≥0⇒y≤13 or y≥3
But A,B>0 and A+B=π/3
⇒ A,B<π/3
⇒ tanAtanB<3 so, y≥3 is not a possibility.
Therefore, y≤13 i.e., max. value of y is 1/3,