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

$\frac{1}{3}$

Explanation for correct option:

Step1. Finding value of $A\&B$

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

Since $A>0,B>0$ therefore $\tan \left(A\right)>0,\tan \left(B\right)>0$

Also, we know that

$\mathit{A}\mathit{M}\mathbf{}\mathbf{\ge }\mathbf{}\mathit{G}\mathit{M}$

therefore,

$\begin{array}{rcl}\frac{\tan \left(A\right)+\tan \left(B\right)}{2}& \ge & \surd \left(\tan \right(A\left)\tan \right(B)\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\tan \left(A\right)+\tan \left(B\right)& \ge & 2\surd \left(\tan \right(A\left)\tan \right(B)\end{array}$

$\Rightarrow$$\begin{array}{rcl}\tan \left(A\right)+\tan \left(B\right)-2\surd \left(\tan \right(A\left)\tan \right(B)& =& 0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{\left[\sqrt{\tan (A}\right)–\sqrt{\tan \left(B\right)}]}^2& =& 0\end{array}$

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

$\Rightarrow$ $\tan \left(A\right)=\tan \left(B\right)$

$\Rightarrow$ $A=B$

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

$\Rightarrow$ $\begin{array}{rcl}A+A& =& \frac{\pi }{3}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}A& =& \frac{\pi }{6}\\ & =& B\end{array}$

Step2. Find the maximum value of $\tan \left(A\right)\tan \left(B\right)$

The maximum value of,$\begin{array}{l}\tan \left(A\right)\tan \left(B\right)\end{array}$

$\begin{array}{rcl}& =& \tan \frac{\pi }{6}\tan \frac{\pi }{6}\\ & =& \left(\frac{1}{\sqrt{3}}\right)\left(\frac{1}{\sqrt{3}}\right)\\ & =& \frac{\mathbf{1}}{\mathbf{3}}\end{array}$

Hence, the correct option is $\left(B\right)$