If $A+B=\pi/3$ then what will maximum value of $\tan(A).\tan(B)$? Suppose I am given that $$A+B=\frac{\pi}{3}$$ then what will be maximum value of $$\tan(A).\tan(B)=?$$
$$\tan(A+B)=\frac{\tan(A)+\tan(B)}{1-\tan(A).\tan(B)}=\sqrt{3}$$
then 
$$\tan(A).\tan(B)=\frac{\sqrt(3)-[\tan(A)+\tan(B)]}{\sqrt{3}}=\lambda$$
now for $\lambda$ to be maximum $\tan(A)+\tan(B)$ should be minimum , how to minimise it? I can make guesses but that's not a good approach.
 A: We have
$$\tan A\tan B=\tan A\tan(\pi/3-A)=\tan A\frac{\tan(\pi/3)-\tan A}{1+\tan(\pi/3)\tan A}=x\frac{\sqrt3-x}{1+x\sqrt 3}=:f(x)$$
Now by differentiating $f$ we can study the variations of $f$.
A: There have to be some restrictions on $A$ and $B$; otherwise $\tan A\>\tan B$ would be unbounded. I'm going to impose the conditions $A\geq0$, $B\geq0$. Put
$$A:={\pi\over6}+\tau,\quad B:={\pi\over6}-\tau\qquad\bigl(|\tau|\leq{\pi\over6}\bigr)\ .$$
Then $A+B={\pi\over3}$, and 
$$\tan A\>\tan B={{1\over\sqrt{3}}+\tan\tau\over 1-{1\over\sqrt{3}}\tan\tau}\cdot{{1\over\sqrt{3}}-\tan\tau\over 1+{1\over\sqrt{3}}\tan\tau}={1-3\tan^2\tau\over3-\tan^2\tau}=3-{8\over3-\tan^2\tau}\ .$$
It follows that the quantity in question is maximal $(={1\over3})$ when $\tau=0$, i.e., when $A=B={\pi\over6}$.
A: set
$$
F(A,B)=\tan A \tan B - \lambda(A+B-\frac \pi 3)
$$
so for an extremum we require:
$$
\sec^2 A \tan B = \lambda \\
\tan A \sec^2 B = \lambda
$$
if $A$ and $B$ are positive angles we may assume that $\cos A \cos B \ne 0$ so the condition reduces to
$$
\sin A \cos A = \sin B \cos B
$$
so $A=B=\frac \pi 6$ and the corresponding value of $\tan A \tan B$ is $\frac13$
A: From Jensen's inequality, the average of the tangent is $\ge$ the tangent of the average, with equality only when $A=B$.
