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