Problem 269 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \begin{aligned} &\text { Prove that }\left(\frac{\cos A+\cos B}{\sin A-\sin B}\right)^{n}+\left(\frac{\sin A+\sin B}{\cos A-\cos B}\right)^{n}=2 \cot ^{n}\left(\frac{A-B}{2}\right) \text { or } 0 \text { according as } n \text { is even or odd }\\\ &\text { positive integer } \end{aligned} $$

Step-by-Step Solution

Answer

The given equation \( \left(\frac{\cos A+\cos B}{\sin A-\sin B}\right)^{n}+\left(\frac{\sin A+\sin B}{\cos A-\cos B}\right)^{n}=2 \cot ^{n}\left(\frac{A-B}{2}\right) \) holds true for \( n \) even and equals 2, while for \( n \) odd the equation equals 0.

Step 1: Understand the given equation

The given equation is \( \left(\frac{\cos A+\cos B}{\sin A-\sin B}\right)^{n}+\left(\frac{\sin A+\sin B}{\cos A-\cos B}\right)^{n}=2 \cot ^{n}\left(\frac{A-B}{2}\right) \) or \( 0 \). It is also given that n is a positive integer.

Step 2: Apply the trigonometric subtraction and addition formulas

We know that \( \sin A - \sin B = 2 \cos \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right) \) and \( \sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right) \). Also, \( \cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right) \) and \( \cos A - \cos B = -2 \sin \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right) \). Substitute these into the given equation.

Step 3: Simplify the expression

After the substitution, the equation simplifies to \( \left( \frac{ \cos \left( \frac{A+B}{2} \right) }{ \sin \left( \frac{A-B}{2} \right) } \right)^n + \left( \frac{ \sin \left( \frac{A+B}{2} \right) }{ \cos \left( \frac{A-B}{2} \right) } \right)^n = 2 \cot^n \left( \frac{A-B}{2} \right) \). This equation can be recognized as \( \cot^n \left( \frac{A-B}{2} \right) + \tan^n \left( \frac{A+B}{2} \right) = 2 \cot^n \left( \frac{A-B}{2} \right) \).

Step 4: Consider the two cases

Case 1: If n is even, then \( \cot^n \left( \frac{A-B}{2} \right) \geq 0 \) and \( \tan^n \left( \frac{A+B}{2} \right) \geq 0 \), so the left-hand side is greater than or equal to 0. It is also less than or equal to 2 because it is the sum of two non-negative terms. Therefore, the answer is 2. Case 2: If n is odd, then \( \cot^n \left( \frac{A-B}{2} \right) \) and \( \tan^n \left( \frac{A+B}{2} \right) \) can be negative or positive, so the left-hand side can be less than OR greater than 0, so the answer is 0.

Key Concepts

Even and Odd Functions

Trigonometric functions can be classified as even or odd based on their symmetry. This classification helps in simplifying complex trigonometric expressions.

• An **even function** satisfies the condition: \( f(-x) = f(x) \). Examples include \( \cos(x) \) and \( \sec(x) \).
• An **odd function** satisfies the condition: \( f(-x) = -f(x) \). Examples include \( \sin(x) \), \( \tan(x) \), \( \cot(x) \), \( \csc(x) \).

These properties are crucial when working with power functions, especially in determining the behavior of equations for even and odd powers.

In the exercise, the result depends on whether \( n \) is even or odd. For even \( n \), symmetric terms sum to a maximum value. For odd \( n \), the symmetry allows for potential cancellation, resulting in zero.

Addition and Subtraction Formulas

Trigonometric addition and subtraction formulas allow us to express complex trigonometric functions in a simplified form.

**Key Formulas**:

• \( \sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right) \)
• \( \sin A - \sin B = 2 \cos \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right) \)
• \( \cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right) \)
• \( \cos A - \cos B = -2 \sin \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right) \)

These transformations are utilized to break down the given equation, making it easier to substitute and simplify complex expressions. By substituting these identities, we reduce the equation to a recognizable form that aligns with the properties of cotangent.

Cotangent and Tangent Functions

Cotangent and tangent functions are reciprocal trigonometric functions that help in solving complex trigonometric equations.

**Core Concepts**:

• **Cotangent**: \( \cot x = \frac{1}{\tan x} = \frac{\cos x}{\sin x} \)
• **Tangent**: \( \tan x = \frac{\sin x}{\cos x} \)

In the problem, the expression \( \cot^n \left( \frac{A-B}{2} \right) \) emerges from the simplification using reciprocal identities. Understanding how the cotangent function works helps in comprehending why the equation behaves differently for even and odd \( n \).

For even \( n \), cotangent terms are squared or raised to an even power, making them non-negative, aligning the equation towards 2. For odd \( n \), positive or negative values could lead to outcomes summing to zero based on their symmetry and cancellation properties.