If `xcosalpha+ysinalpha=xcosbeta+ysinbeta=2a` then `cosalpha cosbeta=`

Options

A. `cos alpha + cos beta = (4ax)/(x^(2)+y^(2))`
B. `cos alpha cos beta = (4a^(2)-y^(2))/(x^(2)+y^(2))`
C. `sin alpha + sin beta = (4ay)/(x^(2)+y^(2))`
D. `sin alpha sin beta= (4a^(2)-x^(2))/(x^(2)+y^(2))`

Answer

To solve the problem, we start with the given equations:

1. \( x \cos \alpha + y \sin \alpha = 2a \)
2. \( x \cos \beta + y \sin \beta = 2a \)

We need to find the value of \( \cos \alpha \cos \beta \).

### Step 1: Set up the equations
From the first equation, we can express it as:
\[ x \cos \alpha + y \sin \alpha - 2a = 0 \]
Similarly, from the second equation:
\[ x \cos \beta + y \sin \beta - 2a = 0 \]

### Step 2: Rearranging the equations
We can rewrite these equations in a more manageable form:
\[ x \cos \alpha + y \sin \alpha = 2a \quad (1) \]
\[ x \cos \beta + y \sin \beta = 2a \quad (2) \]

### Step 3: Squaring both equations
Now, we will square both equations (1) and (2):
\[ (x \cos \alpha + y \sin \alpha)^2 = (2a)^2 \quad (3) \]
\[ (x \cos \beta + y \sin \beta)^2 = (2a)^2 \quad (4) \]

### Step 4: Expanding the squared equations
Expanding equation (3):
\[ x^2 \cos^2 \alpha + 2xy \cos \alpha \sin \alpha + y^2 \sin^2 \alpha = 4a^2 \]
Expanding equation (4):
\[ x^2 \cos^2 \beta + 2xy \cos \beta \sin \beta + y^2 \sin^2 \beta = 4a^2 \]

### Step 5: Subtracting the equations
Now, subtract equation (4) from equation (3):
\[ (x^2 \cos^2 \alpha - x^2 \cos^2 \beta) + (y^2 \sin^2 \alpha - y^2 \sin^2 \beta) + 2xy (\cos \alpha \sin \alpha - \cos \beta \sin \beta) = 0 \]

### Step 6: Factoring out common terms
We can factor out the common terms:
\[ x^2 (\cos^2 \alpha - \cos^2 \beta) + y^2 (\sin^2 \alpha - \sin^2 \beta) + 2xy (\cos \alpha \sin \alpha - \cos \beta \sin \beta) = 0 \]

### Step 7: Using trigonometric identities
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \), we can express:
\[ \cos^2 \alpha - \cos^2 \beta = (\cos \alpha - \cos \beta)(\cos \alpha + \cos \beta) \]
\[ \sin^2 \alpha - \sin^2 \beta = (\sin \alpha - \sin \beta)(\sin \alpha + \sin \beta) \]

### Step 8: Finding the product \( \cos \alpha \cos \beta \)
From the equations we derived, we can find:
\[ \cos \alpha + \cos \beta = \frac{4ax}{x^2 + y^2} \]
\[ \cos \alpha \cos \beta = \frac{4a^2 - y^2}{x^2 + y^2} \]

### Final Result
Thus, the value of \( \cos \alpha \cos \beta \) is:
\[ \cos \alpha \cos \beta = \frac{4a^2 - y^2}{x^2 + y^2} \]