`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(a-x)`
D. `cos alpha+cos beta=2cos alphacos beta.`

Answer

To solve the problem step by step, we start with the given equations and conditions:

### Given:
1. \( x \cos \alpha + y \sin \alpha = 2a \) (Equation 1)
2. \( x \cos \beta + y \sin \beta = 2a \) (Equation 2)
3. \( 2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2} = 1 \)

### Step 1: Rearranging Equations
From Equation 1, we can express \( y \sin \alpha \) in terms of \( x \cos \alpha \) and \( a \):
\[ y \sin \alpha = 2a - x \cos \alpha \]
From Equation 2, we can express \( y \sin \beta \):
\[ y \sin \beta = 2a - x \cos \beta \]