If $A>0,B>0\text{ and }A+B=\frac{\pi }{6}$then the minimum value of $\tan A+\tan B$ is

Solution:

We know that, $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

$\begin{array}{l}\Rightarrow \frac{1}{\sqrt{3}}=\frac{y}{1-\tan A\tan B},\text{ where }y=\tan A+\tan B\\ \Rightarrow \tan A\tan B=1-\sqrt{3}y\end{array}$

Also $\text{ A.M. }\ge \text{ G.M. }$

$\begin{array}{l}\Rightarrow \frac{\tan A+\tan B}{2}\ge \sqrt{\tan A\tan B}\\ \Rightarrow y\ge 2\sqrt{1-\sqrt{3}y}\Rightarrow y^2\ge 4-4\sqrt{3}y\Rightarrow y^2+4\sqrt{3}y-4\ge 0\\ \Rightarrow y\le -2\sqrt{3}-4\text{ or }y\ge -2\sqrt{3}+4\end{array}$