Question

If A > 0, B > 0 and $A + B = \dfrac{\pi }{3}$, then the maximum value of $\tan A\tan B$ is
A. $\dfrac{1}{{\sqrt 3 }}$
B. $\dfrac{1}{3}$
C. 3
D. $\sqrt 3$

Answer 1

Here, we will use the formula for tangent of sum of two angles, and the relation between arithmetic and geometric mean to get the inequality. We will first find the range of A and B. then we will find the range for the tangent of both the angles. We will then find the range for the product of the tangent of both the angles. We will substitute the values in the formula and then find the arithmetic and geometric mean of the two numbers. This will give us inequality and using the inequality obtained we will find the maximum value.

Formula used:
The tangent of sum of two angles is given by $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$. The arithmetic mean and geometric mean of two numbers $a$ and $b$ are given by $\dfrac{{a + b}}{2}$ and $\sqrt {ab}$ respectively, where $A.M. \geqslant G.M.$.

Complete step by step solution:
The angles $A$ and $B$ are positive, and their sum is $\dfrac{\pi }{3}$.
Therefore, we get
$0 < A < \dfrac{\pi }{3}$ and $0 < B < \dfrac{\pi }{3}$
Using tangents, we get
$\Rightarrow \tan 0 < \tan A < \tan \dfrac{\pi }{3}$ and $\tan 0 < \tan B < \tan \dfrac{\pi }{3}$
$\Rightarrow 0 < \tan A < \sqrt 3$ and $0 < \tan B < \sqrt 3$
From the above, we can evaluate that

\Rightarrow \tan A\tan B < 3 \ldots \ldots \ldots \left( 1 \right) \\\\$$ Therefore, the product of the tangents of the angles $$A$$ and $$B$$ should be less than 3. Now, we will use the formula for the tangent of the sum of the angles $$A$$ and $$B$$. Thus, we get $$\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$$ Substituting $$A + B = \dfrac{\pi }{3}$$, we get $$\Rightarrow \tan \dfrac{\pi }{3} = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}} \\\ \Rightarrow \sqrt 3 = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}} \\\ \Rightarrow \sqrt 3 \left( {1 - \tan A\tan B} \right) = \tan A + \tan B \\\ \Rightarrow \tan A + \tan B = \sqrt 3 \left( {1 - \tan A\tan B} \right) \ldots \ldots \ldots \left( 2 \right) \\\\$$ We will find the arithmetic and geometric mean of the two numbers $$\tan A$$ and $$\tan B$$. The arithmetic mean and geometric mean of two numbers $$a$$ and $$b$$ are given by $$\dfrac{{a + b}}{2}$$ and $$\sqrt {ab} $$ respectively, where $$A.M. \geqslant G.M.$$. Therefore, we get $$A.M. = \dfrac{{\tan A + \tan B}}{2}$$ and $$G.M. = \sqrt {\tan A\tan B} $$ Substituting the values of the arithmetic and geometric mean in the relation $$A.M. \geqslant G.M.$$, we get $$\dfrac{{\tan A + \tan B}}{2} \geqslant \sqrt {\tan A\tan B} $$ Simplifying the expression, we get $$ \Rightarrow \tan A + \tan B \geqslant 2\sqrt {\tan A\tan B} $$ Substituting $$\tan A + \tan B = \sqrt 3 \left( {1 - \tan A\tan B} \right)$$ from equation $$\left( 2 \right)$$, we get $$ \Rightarrow \sqrt 3 \left( {1 - \tan A\tan B} \right) \geqslant 2\sqrt {\tan A\tan B} $$ Squaring both sides of the inequation, we get $$ \Rightarrow 3{\left( {1 - \tan A\tan B} \right)^2} \geqslant 4\tan A\tan B$$ Let us assume that $$p = \tan A\tan B$$ to simplify the calculations. Thus, the inequation becomes $$ \Rightarrow 3{\left( {1 - p} \right)^2} \geqslant 4p$$ Applying the algebraic identity $${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$$, we get $$\Rightarrow 3\left( {1 - 2p + {p^2}} \right) \geqslant 4p \\\ \Rightarrow 3 - 6p + 3{p^2} \geqslant 4p \\\\$$ Subtracting $$4p$$ from both sides, we get $$ \Rightarrow 3{p^2} - 10p + 3 \geqslant 0$$ Factorise the expression on the left side of the inequation, we get $$\Rightarrow 3{p^2} - 9p - p + 3 \geqslant 0 \\\ \Rightarrow 3p\left( {p - 3} \right) - 1\left( {p - 3} \right) \geqslant 0 \\\ \Rightarrow \left( {3p - 1} \right)\left( {p - 3} \right) \geqslant 0 \\\ \Rightarrow 3\left( {p - \dfrac{1}{3}} \right)\left( {p - 3} \right) \geqslant 0 \\\ \Rightarrow \left( {p - \dfrac{1}{3}} \right)\left( {p - 3} \right) \geqslant 0 \\\\$$ Now, we know that if $$\left( {x - a} \right)\left( {x - b} \right) \geqslant 0$$ where $$a < b$$, then either $$x \leqslant a$$ or $$x \geqslant b$$. Therefore, from the inequation $$\left( {p - \dfrac{1}{3}} \right)\left( {p - 3} \right) \geqslant 0$$, we get $$p \leqslant \dfrac{1}{3}$$ or $$p \geqslant 3$$. Substituting $$p = \tan A\tan B$$, we get $$\tan A\tan B \leqslant \dfrac{1}{3}$$ or $$\tan A\tan B \geqslant 3$$. However, from inequation $$\left( 1 \right)$$, we know that $$\tan A\tan B < 3$$. Therefore, $$\tan A\tan B \geqslant 3$$ is not possible. $$\therefore \tan A\tan B \leqslant \dfrac{1}{3}$$ Hence, the maximum value of $$\tan A\tan B$$ is $$\dfrac{1}{3}$$. The correct option is option (b). **Note:** A common mistake in this question is to leave the answer at $$\tan A\tan B < 3$$. We should remember that this is not the maximum value of $$\tan A\tan B$$, because the $$ < $$ sign is used, not the $$ \leqslant $$ sign.