Problem 26 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

If \(A>0\) and \(B>0\) and \(A+B=\frac{\pi}{3}\), then the maximum value of \(\tan A \cdot \tan B\) is (a) \(\frac{1}{2}\) (b) \(\frac{1}{3}\) (c) \(\frac{1}{\sqrt{3}}\) (d) \(\frac{1}{\sqrt{2}}\)

Step-by-Step Solution

Answer

The maximum value of \(\tan A \cdot \tan B\) is \(\frac{1}{3}\). Thus, the correct answer is (b) \(\frac{1}{3}\).

Step 1: Expressing B in terms of A

Since the problem states that \(A+B=\frac{\pi}{3}\), this equation can be rearranged to express B in terms of A: \(B=\frac{\pi}{3}-A\).

Step 2: Expressing the target term in terms of A

By substituting \(B=\frac{\pi}{3}-A\) into \(\tan A \cdot \tan B\), the expression becomes \(\tan A \cdot \tan (\frac{\pi}{3}-A)\).

Step 3: Using a trigonometric identity

Applying the tangent formula for difference of angles: \(\tan(\alpha-\beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}\), where \(\alpha=\frac{\pi}{3}\) and \(\beta=A\), the expression \(\tan(\frac{\pi}{3} - A)\) simplifies to \(\frac{\sqrt{3} - \tan A}{1 + \sqrt{3} \tan A}\).

Step 4: Simplifying the target term

Substituting the simplified expression of \(\tan(\frac{\pi}{3}-A)\) from step 3 into \(\tan A \cdot \tan (\frac{\pi}{3}-A)\) results in the expression \(\tan A \cdot \frac{\sqrt{3} - \tan A}{1 + \sqrt{3} \tan A}\).

Step 5: Finding the derivative

Differentiating the simplified expression with respect to A and setting equal to zero gives us the maximum value of the expression.

Step 6: Solving

Solving for A, we find that the maximum value of \(\tan A \cdot \tan B\) is achieved when \(A = \frac{\pi}{6}\). Substituting \(A = \frac{\pi}{6}\) back into the expression gives a maximum value of \(\frac{1}{3}\).

Key Concepts

Trigonometric Identities

Understanding trigonometric identities is essential for solving complex trigonometry problems. These identities are equations involving trigonometric functions that are true for every value of the variable involved. They allow us to simplify trigonometric expressions, transform them into different forms, and solve for unknown angles and lengths.

One of the fundamental identities used in trigonometry is the tangent identity for the difference of angles. This identity is given by:
\[\begin{equation}\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}\end{equation}\]
Another set of basic trigonometric identities includes the Pythagorean identity for the tangent function:\[\begin{equation}\tan^2 A + 1 = \sec^2 A\end{equation}\]Understanding these identities allows for the conversion of the original expression involving the product of tangents into a simpler form that is easier to manipulate. This simplification is the key to finding the maximum value in the given exercise.

For the given problem, knowing these identities permits the conversion of \(\tan A \cdot \tan B\) into a form that clearly shows how A and B are connected, eventually leading to the determination of the maximum value.

Tangent of the Sum and Difference of Angles

The tangent of the sum and difference of angles formula is an invaluable tool for trigonometric problem-solving. This formula expresses the tangent of an angle created by either the sum or difference of two other angles. Here's what this looks like for the difference of two angles:
\[\begin{equation}\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}\end{equation}\]
In the exercise, we applied this identity to express \(\tan(\frac{\pi}{3} - A)\) as \(\frac{\sqrt{3} - \tan A}{1 + \sqrt{3} \tan A}\), which plays a fundamental role in manipulating the expression to find the maximum value. The formula for the sum of angles, although not used in this exercise, is similarly structured and equally important in other contexts.

The clever use of these formulas often leads to significant simplifications that unmask the underlying structure of the problem, ultimately leading to its resolution. When A and B sum to a constant, such as \(\frac{\pi}{3}\), knowing these formulas allows us to establish a single-variable function that can be optimized to find maximum or minimum values.

Finding Maximum Value in Trigonometry

Finding the maximum value of a trigonometric expression is a common task in trigonometry, often requiring knowledge of calculus. When the expression is reduced to a single variable function, as we did by setting \(B = \frac{\pi}{3} - A\) and using trigonometric identities, we can then use calculus to locate maximum values. The process usually involves taking the derivative of the function with respect to the variable, setting it equal to zero, and solving for the variable.

This procedure leads to the identification of critical points—values of the variable where the function's slope is zero. If a critical point is a maxima, the function has a maximum value at that point. In the context of our exercise, by deriving the function of \(\tan A\) after simplifying \(\tan A \cdot \tan B\) and setting the derivative equal to zero, we can identify the angle A at which the product of tangents attains its maximum value.

Knowing how to apply differentiation to maximize or minimize a function is an incredibly powerful technique that goes beyond just trigonometry, having applications across all fields of calculus and optimization problems.