Question

lf $A>0,B>0$, 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. $3$
• D. $\sqrt{3}$

Answer 1

Answer: (B)

$\frac{1}{3}$

$\tan A\tan (\frac{\pi }{3}-A)=\frac{\tan A(\sqrt{3}-\tan A)}{1+\sqrt{3}\tan A}$
$f\left(A\right)=\frac{\sqrt{3}\tan A-{\tan }^{2}A}{1+\sqrt{3}\tan A}$
$f^{\prime }\left(A\right)=se{c}^{2}A(\frac{(\sqrt{3}-2\tan A)(1-\sqrt{3}\tan A)-\sqrt{3}(\sqrt{3}\tan A-{\tan }^{2}A)}{{(1+\sqrt{3}\tan A)}^2}$
$f^{\prime }\left(A\right)={\sec }^{2}A\frac{(\sqrt{3}+\tan A-\sqrt{3}{\tan }^{2}A-3\tan A)}{(1-\sqrt{3}\tan A)}$
So $f^{\prime }\left(A\right)=0$ when $\tan A=\frac{1}{\sqrt{3}},\sqrt{3}$

Check for $f^{\prime \prime }\left(A\right)<0$
$f\left(A\right)$ is maximum when $\tan A=\frac{1}{\sqrt{3}}$
$f\left(A\right)=\frac{1}{\sqrt{3}}\times \frac{1}{\sqrt{3}}=\frac{1}{3}$