Question

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

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{6}$

Answer 1

The correct option is B

$\frac{1}{3}$

Finding the value for :

Given, $A>0,B>0andA+B=\frac{\mathrm{\pi }}{3}$

That means,$\tan \left(A\right)>0,\tan \left(B\right)>0$

We know that, $\mathit{A}\mathit{M}\mathbf{\ge }\mathit{G}\mathit{M}$

$\frac{(\tan \left(\mathrm{A}\right)+\tan \left(\mathrm{B}\right))}{2}\ge \sqrt{\tan \left(\mathrm{A}\right)\tan \left(\mathrm{B}\right)}$

$\Rightarrow$$(\tan \left(\mathrm{A}\right)+\tan \left(\mathrm{B}\right))\ge 2\sqrt{\tan \left(\mathrm{A}\right)\tan \left(\mathrm{B}\right)}$

$\Rightarrow$$(\tan \left(\mathrm{A}\right)+\tan \left(\mathrm{B}\right))-2\sqrt{\tan \left(\mathrm{A}\right)\tan \left(\mathrm{B}\right)}=0$

$\Rightarrow$${(\sqrt{\tan \mathrm{A}}-\sqrt{\tan \left(\mathrm{B}\right)})}^2=0$

$\Rightarrow$$\sqrt{\tan \left(\mathrm{A}\right)}-\sqrt{\tan \left(\mathrm{B}\right)}=0$

$\Rightarrow$ $$\begin{gathered} \sqrt{\tan \left(\mathrm{A}\right)}=\sqrt{\tan \left(\mathrm{B}\right)} \\ tan\left(\mathrm{A}\right)=tan\left(\mathrm{B}\right) \end{gathered}$$

According to the given, this is possible when $A=B$

$=\frac{\mathrm{\pi }}{6}$

Therefore, $\tan \left(A\right)\tan \left(B\right)$$=\tan \left(\frac{\pi }{6}\right)\tan \left(\frac{\pi }{6}\right)$

$\Rightarrow$ $=\left(\frac{1}{\sqrt{3}}\right)\left(\frac{1}{\sqrt{3}}\right)$

$\Rightarrow$$={\left(\frac{1}{\sqrt{3}}\right)}^2$

$\Rightarrow$ $\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}$

Hence, option 'B' is correct.