Question

If $A>0,B>0$ and $A+B=\frac{\pi }{3},$ then the maximum value of $3\tan A\tan B$ is

Answer 1

Let $y=\tan A\tan B,$
So $A,B>0$ and $A+B=\frac{\pi }{3}\Rightarrow A,B<\frac{\pi }{3}$
$\Rightarrow \tan A\tan B<3\Rightarrow y\in (0,3)\cdots \left(i\right)$
Now
$A+B=\frac{\pi }{3}\Rightarrow \tan (A+B)=\sqrt{3}$
$\Rightarrow \frac{\tan A+\tan B}{1-\tan A\tan B}=\sqrt{3}$
$\Rightarrow \frac{\tan A+\frac{y}{\tan A}}{1-y}=\sqrt{3}$
$\Rightarrow {\tan }^{2}A+\sqrt{3}(y-1)\tan A+y=0$
For real +ive value of $\tan A$,
$3(y-1{)}^2-4y\ge 0$
$\Rightarrow 3y^2-10y+3\ge 0$
$\Rightarrow (y-3)(y-\frac{1}{3})\ge 0$
$y\in (-\infty ,\frac{1}{3}]\cup [3,\infty )\cdots \left(ii\right)$
$\therefore$ from $\left(i\right)$ and $\left(ii\right)$ $y\in (0,\frac{1}{3}]$
So, maximum value of $3y$ is $1.$