If `(cosA)/(cosB)=(x)/(y)`, where `A!=B`, then

Options

A. `(xtanA+ytanB)/(x+y)=tan((A+B)/(2))`
B. `(xtanA-ytanB)/(x+y)=tan((A-B)/(2))`
C. `(ysinA+xsinB)/(ysinA-xsinB)=(sin(A+B))/(sin(A-B))`
D. `xcosA+ycosB=0`

Answer

To solve the problem given that \(\frac{\cos A}{\cos B} = \frac{x}{y}\) where \(A \neq B\), we need to check the validity of the four options provided.

### Step-by-Step Solution

#### Option 1:
We need to verify if

\[ \frac{x \tan A + y \tan B}{x + y} = \tan\left(\frac{A + B}{2}\right) \]

1. Start with the left-hand side (LHS):

\[ \text{LHS} = \frac{x \tan A + y \tan B}{x + y} \]

2. Substitute \(\tan A = \frac{\sin A}{\cos A}\) and \(\tan B = \frac{\sin B}{\cos B}\):

\[ = \frac{x \frac{\sin A}{\cos A} + y \frac{\sin B}{\cos B}}{x + y} \]

3. Now substitute \(x = k \cos A\) and \(y = k \cos B\) (where \(k\) is a constant):

\[ = \frac{k \cos A \frac{\sin A}{\cos A} + k \cos B \frac{\sin B}{\cos B}}{k \cos A + k \cos B} \]

4. Simplifying gives:

\[ = \frac{k \sin A + k \sin B}{k (\cos A + \cos B)} = \frac{\sin A + \sin B}{\cos A + \cos B} \]

5. Using the sine addition formula, we can express this as:

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

Thus, Option 1 is **correct**.

#### Option 2:
We need to verify if

\[ \frac{x \tan A - y \tan B}{x + y} = \tan\left(\frac{A - B}{2}\right) \]

1. Start with the left-hand side (LHS):

\[ \text{LHS} = \frac{x \tan A - y \tan B}{x + y} \]

2. Substitute \(\tan A\) and \(\tan B\) as before:

\[ = \frac{x \frac{\sin A}{\cos A} - y \frac{\sin B}{\cos B}}{x + y} \]

3. Again substitute \(x = k \cos A\) and \(y = k \cos B\):

\[ = \frac{k \sin A - k \sin B}{k (\cos A + \cos B)} = \frac{\sin A - \sin B}{\cos A + \cos B} \]

4. Using the sine subtraction formula:

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

Thus, Option 2 is **correct**.

#### Option 3:
We need to verify if

\[ \frac{y \sin A + x \sin B}{y \sin A - x \sin B} = \frac{\sin(A + B)}{\sin(A - B)} \]

1. Start with the left-hand side (LHS):

\[ \text{LHS} = \frac{y \sin A + x \sin B}{y \sin A - x \sin B} \]

2. Substitute \(x = k \cos A\) and \(y = k \cos B\):

\[ = \frac{k \cos B \sin A + k \cos A \sin B}{k \cos B \sin A - k \cos A \sin B} \]

3. This simplifies to:

\[ = \frac{\cos B \sin A + \cos A \sin B}{\cos B \sin A - \cos A \sin B} \]

4. Using the sine addition and subtraction formulas:

\[ = \frac{\sin(A + B)}{\sin(A - B)} \]

Thus, Option 3 is **correct**.

#### Option 4:
We need to verify if

\[ x \cos A + y \cos B = 0 \]

1. Substitute \(x = k \cos A\) and \(y = k \cos B\):

\[ k \cos A \cos A + k \cos B \cos B = 0 \]

2. This simplifies to:

\[ k(\cos^2 A + \cos^2 B) = 0 \]

Since \(k\) cannot be zero, this equation does not hold true. Thus, Option 4 is **incorrect**.

### Final Conclusion
- **Option 1**: Correct
- **Option 2**: Correct
- **Option 3**: Correct
- **Option 4**: Incorrect