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. $3$
• D. $\sqrt{3}$

Answer 1

The correct option is B $\frac{1}{3}$

The given equation is:

$A+B=\frac{\pi }{3}$

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

$\Rightarrow f\left(A\right)=\tan A\tan B=1-\sqrt{3}(\tan A+\tan (\frac{\pi }{3}-A))$

To find the extremum points we differentiate and equate it to zero

$\Rightarrow f^{\prime }\left(A\right)=-\sqrt{3}({\sec }^{2}A-{\sec }^2(\frac{\pi }{3}-A))$

$f^{\prime }\left(A\right)=0$

$\Rightarrow -\sqrt{3}({\sec }^{2}A-{\sec }^2(\frac{\pi }{3}-A))=0$

$\Rightarrow {\sec }^{2}A={\sec }^2(\frac{\pi }{3}-A))$

$\Rightarrow \cos A=\cos (\frac{\pi }{3}-A))$

$\Rightarrow A=\frac{\pi }{3}-A+2\pi$

$\Rightarrow 2A=\frac{7\pi }{3}$

$\Rightarrow A=\frac{7\pi }{6}$

$\Rightarrow \frac{7\pi }{6}+B=\frac{\pi }{3}$

$\Rightarrow B=-\frac{5\pi }{3}$

Thus sunstituting the value of A and B such that tanAtanB is maximum we get,

$\Rightarrow \tan A\tan B=\tan \frac{7\pi }{6}\tan -\frac{5\pi }{3}$

$\Rightarrow \tan A\tan B=\frac{1}{\sqrt{3}}\times \frac{1}{\sqrt{3}}$

$\Rightarrow \tan A\tan B=\frac{1}{3}$ .....Answer