Question

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

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

Answer 1

The correct option is D $4-2\sqrt{3}$

$A+B=\frac{\pi }{6}\text{Let}y=\tan A+\tan B=\tan A+\tan (\frac{\pi }{6}-A)\frac{dy}{dA}={\sec }^{2}A+{\sec }^2(\frac{\pi }{6}-A)(-1)={\sec }^{2}A-{\sec }^2(\frac{\pi }{6}-A)\frac{dy}{dA}=0\Rightarrow {\sec }^{2}A={\sec }^2(\frac{\pi }{6}-A)\Rightarrow A=\frac{\pi }{6}-A\Rightarrow A=\frac{\pi }{12}=B$
Minimum value of $\tan A+\tan B=2\tan \left(\frac{\pi }{12}\right)=2(2-\sqrt{3})=4-2\sqrt{3}$

OR,
$\text{Since,}A,B>0\&A+B=\frac{\pi }{6}\Rightarrow \tan A,\tan B>0$
Using AM-GM inequality $(AM\ge GM)$
$\frac{\tan A+\tan B}{2}\ge \sqrt{\tan A.\tan B}$
The equality is achieved when $A=B$
$\Rightarrow A=B=\frac{\pi }{12}$
$\therefore$ Minimum value of $\tan A+\tan B=2\tan \left(\frac{\pi }{12}\right)=2(2-\sqrt{3})=4-2\sqrt{3}$