Question

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

• A. $\frac{1}{\sqrt{3}}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $\sqrt{3}$

Answer 1

Answer: (B)

Let $y=\tan A\tan B$
$y=\tan A\tan \left(\frac{\pi }{3}-A\right)\left[\because A+B=\frac{\pi }{3}\right]$
$\Rightarrow \frac{dy}{dA}={\sec }^{2}A\tan \left(\frac{\pi }{3}-A\right)-{\sec }^2\left(\frac{\pi }{3}-A\right)\tan A$
For maxima or minima, put $\frac{dy}{dA}=0$
$\therefore {\sec }^{2}A\tan \left(\frac{\pi }{3}-A\right)-{\sec }^2\left(\frac{\pi }{3}-A\right)\tan A=0$
$\Rightarrow {\sec }^{2}A\tan \left(\frac{\pi }{3}-A\right)=\tan A{\sec }^2\left(\frac{\pi }{3}-A\right)$
$\Rightarrow \left(1+{\tan }^{2}A\right)\tan \left(\frac{\pi }{3}-A\right)=\tan A\left[1+{\tan }^2\left(\frac{\pi }{3}-A\right)\right]$
$\Rightarrow \tan \left(\frac{\pi }{3}-A\right)+{\tan }^{2}A\tan \left(\frac{\pi }{3}-A\right)=\tan A+\tan A{\tan }^2\left(\frac{\pi }{3}-A\right)$
$\Rightarrow \left[\tan \left(\frac{\pi }{3}-A\right)-\tan A\right]$
$\left[1-\tan A\tan \left(\frac{\pi }{3}-A\right)\right]=0$
$\Rightarrow \tan \left(\frac{\pi }{3}-A\right)=\tan A$
$\Rightarrow \frac{\pi }{3}-A=A$
$\Rightarrow 2A=\frac{\pi }{3}$
$\Rightarrow A=\frac{\pi }{6}$
$\therefore A=B=\frac{\pi }{6}$
Maximum value of $\tan A\tan B=\tan \frac{\pi }{6}\tan \frac{\pi }{6}$
$=\frac{1}{\sqrt{3}}\times \frac{1}{\sqrt{3}}=\frac{1}{3}$