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

Given $\mathrm{A}+\mathrm{B}={60}^{\circ }\text{ or }\mathrm{B}={60}^{\circ }-\mathrm{A}$
$\therefore \tan \mathrm{B}=\tan \left({60}^{\circ }-\mathrm{A}\right)=\frac{\sqrt{3}-\tan \mathrm{A}}{1+\sqrt{3}\tan \mathrm{A}}$
$\text{ Now }\mathrm{z}=\tan \mathrm{Atan} \mathrm{B}$
or    $\mathrm{z}=\frac{\mathrm{t}(\sqrt{3}-\mathrm{t})}{1+\sqrt{3}\mathrm{t}}=\frac{\sqrt{3}\mathrm{t}-{\mathrm{t}}^2}{1+\sqrt{3}\mathrm{t}}$
where t = tan A
       $\frac{\mathrm{dz}}{\mathrm{dt}}=-\frac{(\mathrm{t}+\sqrt{3})(\sqrt{3}\mathrm{t}-1)}{(1+\sqrt{3}\mathrm{t}{)}^2}=0$
or     $\mathrm{t}=1/\sqrt{3}$
or     $\mathrm{t}=\tan \mathrm{A}=\tan {30}^{\circ }$
The other value is rejected as both A and B are +ve acute angles.
If $\mathrm{t}<\frac{1}{\sqrt{3}},\frac{\mathrm{dz}}{\mathrm{dt}}$is positive and if $\mathrm{t}>\frac{1}{\sqrt{3}},\frac{\mathrm{dz}}{\mathrm{dt}}$is negative.
Hence, maximum when $\mathrm{t}=\frac{1}{\sqrt{3}}$and maximum value $=\frac{1}{3}.$