Find the value of `((cosA+cosB)/(sinA-sinB))^(n)+((sinA+sinB)/(cosA-cosB))^(n)` (where, n is an even)

Answer

To find the value of the expression

\[ \left(\frac{\cos A + \cos B}{\sin A - \sin B}\right)^n + \left(\frac{\sin A + \sin B}{\cos A - \cos B}\right)^n \]

where \( n \) is an even integer, we can use trigonometric identities to simplify the expression.

### Step 1: Use Trigonometric Identities

We start by applying the sum-to-product identities for cosine and sine:

1. **Cosine Addition Formula**:
\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]

2. **Sine Subtraction Formula**:
\[ \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]

3. **Sine Addition Formula**:
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]

4. **Cosine Subtraction Formula**:
\[ \cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]

### Step 2: Substitute the Identities

Now, we substitute these identities into the original expression:

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

This simplifies to:

\[ \frac{\cos\left(\frac{A-B}{2}\right)}{\sin\left(\frac{A-B}{2}\right)} = \cot\left(\frac{A-B}{2}\right) \]

Similarly, for the second part:

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

This simplifies to:

\[ -\frac{\cos\left(\frac{A-B}{2}\right)}{\sin\left(\frac{A-B}{2}\right)} = -\cot\left(\frac{A-B}{2}\right) \]

### Step 3: Combine the Results

Now, we can rewrite the original expression:

\[ \left(\cot\left(\frac{A-B}{2}\right)\right)^n + \left(-\cot\left(\frac{A-B}{2}\right)\right)^n \]

Since \( n \) is an even integer, we have:

\[ (-\cot\left(\frac{A-B}{2}\right))^n = \cot^n\left(\frac{A-B}{2}\right) \]

Thus, the expression simplifies to:

\[ \cot^n\left(\frac{A-B}{2}\right) + \cot^n\left(\frac{A-B}{2}\right) = 2 \cot^n\left(\frac{A-B}{2}\right) \]

### Final Result

Therefore, the value of the expression is:

\[ \boxed{2 \cot^n\left(\frac{A-B}{2}\right)} \]