If `cosalpha=1/2(x+1/x)` `cosbeta=1/2(y+1/y)` then `cos(alpha-beta)` is equal to

Options

A. `sin(alpha+beta+gamma)=singammaAAgammainR`
B. `cosalphacosbeta=1AAalpha,beta inR`
C. `(cosalpha+cosbeta)^2=4AAalpha,beta in R`
D. `sin (alpha+beta+gamma)=sinalpha+sinbeta+singammaAAa,b,gammainR`

Explanation

To solve the problem where \( \cos \alpha = \frac{1}{2} \left( x + \frac{1}{x} \right) \) and \( \cos \beta = \frac{1}{2} \left( y + \frac{1}{y} \right) \), we need to find \( \cos(\alpha - \beta) \).

### Step 1: Express \( \cos \alpha \) and \( \cos \beta \)
Given:
\[ \cos \alpha = \frac{1}{2} \left( x + \frac{1}{x} \right) = \frac{x^2 + 1}{2x} \]
\[ \cos \beta = \frac{1}{2} \left( y + \frac{1}{y} \right) = \frac{y^2 + 1}{2y} \]

### Step 2: Use the cosine difference formula
The formula for \( \cos(\alpha - \beta) \) is:
\[ \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \]

### Step 3: Calculate \( \cos \alpha \cos \beta \)
Substituting the expressions for \( \cos \alpha \) and \( \cos \beta \):
\[ \cos \alpha \cos \beta = \left( \frac{x^2 + 1}{2x} \right) \left( \frac{y^2 + 1}{2y} \right) = \frac{(x^2 + 1)(y^2 + 1)}{4xy} \]

### Step 4: Find \( \sin \alpha \) and \( \sin \beta \)
To find \( \sin \alpha \) and \( \sin \beta \), we can use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \).

For \( \sin \alpha \):
\[ \sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \left( \frac{x^2 + 1}{2x} \right)^2 \]
Calculating \( \cos^2 \alpha \):
\[ \cos^2 \alpha = \frac{(x^2 + 1)^2}{4x^2} = \frac{x^4 + 2x^2 + 1}{4x^2} \]
Thus,
\[ \sin^2 \alpha = 1 - \frac{x^4 + 2x^2 + 1}{4x^2} = \frac{4x^2 - (x^4 + 2x^2 + 1)}{4x^2} = \frac{4x^2 - x^4 - 2x^2 - 1}{4x^2} = \frac{-x^4 + 2x^2 - 1}{4x^2} \]
So,
\[ \sin \alpha = \sqrt{\frac{-x^4 + 2x^2 - 1}{4x^2}} = \frac{\sqrt{-x^4 + 2x^2 - 1}}{2x} \]

Similarly for \( \sin \beta \):
\[ \sin^2 \beta = 1 - \cos^2 \beta = 1 - \left( \frac{y^2 + 1}{2y} \right)^2 \]
Calculating \( \cos^2 \beta \):
\[ \cos^2 \beta = \frac{(y^2 + 1)^2}{4y^2} = \frac{y^4 + 2y^2 + 1}{4y^2} \]
Thus,
\[ \sin^2 \beta = 1 - \frac{y^4 + 2y^2 + 1}{4y^2} = \frac{4y^2 - (y^4 + 2y^2 + 1)}{4y^2} = \frac{-y^4 + 2y^2 - 1}{4y^2} \]
So,
\[ \sin \beta = \sqrt{\frac{-y^4 + 2y^2 - 1}{4y^2}} = \frac{\sqrt{-y^4 + 2y^2 - 1}}{2y} \]

### Step 5: Substitute \( \sin \alpha \) and \( \sin \beta \) into the formula
Now substituting into the cosine difference formula:
\[ \cos(\alpha - \beta) = \frac{(x^2 + 1)(y^2 + 1)}{4xy} + \left( \frac{\sqrt{-x^4 + 2x^2 - 1}}{2x} \cdot \frac{\sqrt{-y^4 + 2y^2 - 1}}{2y} \right) \]

### Step 6: Final expression
This gives us the final expression for \( \cos(\alpha - \beta) \):
\[ \cos(\alpha - \beta) = \frac{(x^2 + 1)(y^2 + 1)}{4xy} + \frac{\sqrt{(-x^4 + 2x^2 - 1)(-y^4 + 2y^2 - 1)}}{4xy} \]