If `If x cosalpha+y sinalpha=2a, x cos beta+y sinbeta=2aand 2sin""(alpha)/(2)sin""(beta)/(2)=1,` then

Options

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

Answer

To solve the given problem step by step, we will analyze the equations and derive the necessary relationships.

### Step 1: Write down the given equations
We have two equations:
1. \( x \cos \alpha + y \sin \alpha = 2a \) (Equation 1)
2. \( x \cos \beta + y \sin \beta = 2a \) (Equation 2)

### Step 2: Rearranging the equations
From Equation 1, we can express \( y \sin \alpha \):
\[ y \sin \alpha = 2a - x \cos \alpha \]
From Equation 2, we can express \( y \sin \beta \):
\[ y \sin \beta = 2a - x \cos \beta \]

### Step 3: Square both sides of the rearranged equations
Squaring both sides of the equations we derived:
1. \( (y \sin \alpha)^2 = (2a - x \cos \alpha)^2 \)
2. \( (y \sin \beta)^2 = (2a - x \cos \beta)^2 \)

### Step 4: Expand the squared equations
Expanding both equations:
1. \( y^2 \sin^2 \alpha = 4a^2 - 4a x \cos \alpha + x^2 \cos^2 \alpha \)
2. \( y^2 \sin^2 \beta = 4a^2 - 4a x \cos \beta + x^2 \cos^2 \beta \)

### Step 5: Set the equations equal to each other
Since both expressions equal \( y^2 \sin^2 \alpha \) and \( y^2 \sin^2 \beta \), we can set them equal:
\[ 4a^2 - 4a x \cos \alpha + x^2 \cos^2 \alpha = 4a^2 - 4a x \cos \beta + x^2 \cos^2 \beta \]

### Step 6: Simplify the equation
Cancelling \( 4a^2 \) from both sides:
\[ -4a x \cos \alpha + x^2 \cos^2 \alpha = -4a x \cos \beta + x^2 \cos^2 \beta \]
Rearranging gives:
\[ 4a x (\cos \beta - \cos \alpha) = x^2 (\cos^2 \beta - \cos^2 \alpha) \]

### Step 7: Factor the right-hand side
Using the identity \( \cos^2 \beta - \cos^2 \alpha = (\cos \beta - \cos \alpha)(\cos \beta + \cos \alpha) \):
\[ 4a x (\cos \beta - \cos \alpha) = x^2 (\cos \beta - \cos \alpha)(\cos \beta + \cos \alpha) \]

### Step 8: Cancel common factors
Assuming \( \cos \beta \neq \cos \alpha \), we can divide both sides by \( \cos \beta - \cos \alpha \):
\[ 4a x = x^2 (\cos \beta + \cos \alpha) \]

### Step 9: Rearranging gives us a relation
Rearranging gives:
\[ x^2 (\cos \beta + \cos \alpha) - 4a x = 0 \]
Factoring out \( x \):
\[ x (x (\cos \beta + \cos \alpha) - 4a) = 0 \]

### Step 10: Solve for \( x \)
Thus, we have two cases:
1. \( x = 0 \)
2. \( x (\cos \beta + \cos \alpha) = 4a \)

### Step 11: Use the third relation
The third relation given is:
\[ \frac{2 \sin \frac{\alpha}{2}}{2 \sin \frac{\beta}{2}} = 1 \]
Squaring both sides gives:
\[ 4 \sin^2 \frac{\alpha}{2} = 4 \sin^2 \frac{\beta}{2} \]
Thus, \( \sin^2 \frac{\alpha}{2} = \sin^2 \frac{\beta}{2} \), which implies that \( \alpha = \beta \) or \( \alpha + \beta = 180^\circ \).

### Final Result
From the derived equations, we can conclude that:
1. If \( x \neq 0 \), then \( y^2 = 4a(4a - x) \) is a valid relation.