Animated Solution for Mathematics - Trigonometry: If $A>0,B>0$ and $A+B=\pi /3$, then the maximum value of $\tan A\tan B$ is ..........

Unlocking the Maximum

Welcome, future engineers! Today, we are going to dissect a problem that seems simple on the surface but hides a beautiful, rigorous logic beneath.

We are tasked with finding the maximum value of the product $y=\tan A\tan B$, given the constraint $A+B=\frac{\pi }{3}$ where $A,B>0$.

When you see a problem like this, do not just start calculating. Pause and visualize the relationship. We have two variables, $A$ and $B$, but they are shackled together by a sum. This is our golden ticket.

The Power of Reduction

In trigonometry, having two variables is often a burden. We want to simplify our world.

Since $A+B=\frac{\pi }{3}$, we can immediately express $B$ as $B=\frac{\pi }{3}-A$.

Now, our product $y$ becomes a function of a single variable, $A$:

Suddenly, the problem feels much more manageable. We have moved from a two-variable system to a single-variable function. This is the first step in any JEE problem: reduce the complexity.

The Algebraic Bridge

Now, we need to expand $\tan \left(\frac{\pi }{3}-A\right)$. Recall the fundamental identity for the tangent of a difference:

Applying this to our expression, we get:

Since we know $\tan \frac{\pi }{3}=\sqrt{3}$, our equation transforms into:

This looks a bit intimidating, but let's keep our cool. Cross-multiply the denominator to the left side:

Expanding this gives us $y+\sqrt{3}y\tan A=\sqrt{3}\tan A-{\tan }^{2}A$. Rearranging everything to one side, we arrive at a beautiful quadratic equation in terms of $\tan A$:

This is the heart of the problem.

The Discriminant's Wisdom

Here is where the JEE magic happens. We have a quadratic equation in $\tan A$. For $A$ to be a real angle, $\tan A$ must be a real number.

For a quadratic equation $a{x}^2+bx+c=0$ to have real roots, its discriminant $D=b^2-4ac$ must be greater than or equal to zero. Let's apply this.

Here, $a=1$, $b=\sqrt{3}(y-1)$, and $c=y$. Thus:

Expanding this, we get $3(y^2-2y+1)-4y\ge 0$, which simplifies to $3y^2-10y+3\ge 0$. Factoring this quadratic inequality, we find:

This tells us that $y$ must lie in the regions $y\le \frac{1}{3}$ or $y\ge 3$.

The Final Filter

We are almost there! We have two possible regions for $y$, but which one is correct? We must return to our physical constraints.

We know $A,B>0$ and $A+B=\frac{\pi }{3}$. This implies that both $A$ and $B$ must be strictly between $0$ and $\frac{\pi }{3}$.

Consequently, $\tan A<\sqrt{3}$ and $\tan B<\sqrt{3}$. Therefore, their product $y=\tan A\tan B$ must be strictly less than $(\sqrt{3})(\sqrt{3})=3$.

This forces us to reject the region $y\ge 3$. We are left with the only valid condition: $y\le \frac{1}{3}$.

Thus, the maximum value of our product is $\frac{1}{3}$.

Remember, math is not just about formulas; it is about constraints and logic. You have successfully navigated the algebra, the trigonometry, and the logical filtering. Keep this mindset, and you will conquer any problem the JEE throws at you!