If A > 0, B > 0, and A + B = `pi/3` then the maximum value of tan A tan B is

Options

A. `(1)/(sqrt(3))`
B. `1/3`
C. 3
D. `sqrt(3)`

Answer

2

Explanation

Given A+B =`60^(@)` or `B =60^(@)` -A
`therefore tan B=tan 60^(@)-A=(sqrt(3)-tanA)/(1+sqrt(3)tanA)`
Now z= tan A tanB
or `Z= (t sqrt(3-t))/(1+sqrt(3)t)=sqrt(3t-t^(2))/(1+sqrt(3)t)` ltrbgt where t= tan A
`(dz)/(dt)=(t+sqrt(3)sqrt(3t-1))/(1+sqrt(3)t)^(2)=0` ltrbgt or `t=1//sqrt(3)`
or t=`tanA=tan30^(@)`
The other values is rejected both A and B are + ve acute angles it `t lt(1)/sqrt(3),(dz)/(dt)` is positive and if `tgt(1)/sqrt(3),(dz)/(dt)` is negative
Hene maximum when t =`(1)/sqrt(3)` and maximum value =`1/3`