Question

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

• A. $\sqrt{3}-\sqrt{2}$
• B. $4-2\sqrt{3}$
• C. $\frac{2}{\sqrt{3}}$
• D. $2-\sqrt{3}$

Answer 1

Answer: (B)

$4-2\sqrt{3}$

$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{y}{1-\tan A\tan B}$ where $y=\tan A+\tan B$

$\Rightarrow \tan A\tan B=1-\sqrt{3}y$

Also $AM\ge GM$

$\Rightarrow \frac{\tan A+\tan B}{2}\ge \sqrt{\tan A\tan B}$

$\Rightarrow y\ge 2\sqrt{1-\sqrt{3}y}$

$\Rightarrow y^2\ge 4-4\sqrt{3}y$

$\Rightarrow y^2+4\sqrt{3}y-4\ge 0$

$\Rightarrow y\le -2\sqrt{3}-4$ or $y\ge -2\sqrt{3}+4$

$(y\le -2\sqrt{3}-4$ is not possible as $\tan A;\tan B>0)$