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Question

If A>0,B>0 and A+B=π3, then the maximum value of 3tanAtanB is

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Solution

Let y=tanAtanB,
So A,B>0 and A+B=π3A,B<π3
tanAtanB<3y(0,3)(i)
Now
A+B=π3tan(A+B)=3
tanA+tanB1tanAtanB=3
tanA+ytanA1y=3
tan2A+3(y1)tanA+y=0
For real +ive value of tanA,
3(y1)24y0
3y210y+30
(y3)(y13)0
y(,13][3,)(ii)
from (i) and (ii) y(0,13]
So, maximum value of 3y is 1.

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