If ` 2 cos theta = x + (1)/( x) , 2 cos phi = y + (1)/( y)` then

Options

A. `(x)/(y)+ (y)/(x) = 2 cos (theta - phi)`
B. ` xy + (1)/( xy) = 2 cos ( theta + phi)`
C. ` x^(m) y^(n) + (1)/( x^(m) y^(n)) = 2 cos ( m theta + n phi)`
D. ` (x^(m))/(y^(n)) + (y^(n))/(x^(m)) = 2 cos ( m theta - n phi)`

Answer

To solve the problem given the equations \( 2 \cos \theta = x + \frac{1}{x} \) and \( 2 \cos \phi = y + \frac{1}{y} \), we will derive the values of \( x \) and \( y \) in terms of complex exponentials and then analyze the relationships between them.

### Step 1: Rearranging the equations

Starting with the first equation:
\[ 2 \cos \theta = x + \frac{1}{x} \]
We can multiply both sides by \( x \) to eliminate the fraction:
\[ 2 \cos \theta \cdot x = x^2 + 1 \]
Rearranging gives us a quadratic equation:
\[ x^2 - 2 \cos \theta \cdot x + 1 = 0 \]

### Step 2: Solving the quadratic equation for \( x \)

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1 \), \( b = -2 \cos \theta \), and \( c = 1 \).

Substituting these values into the formula:
\[ x = \frac{2 \cos \theta \pm \sqrt{(2 \cos \theta)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
This simplifies to:
\[ x = \frac{2 \cos \theta \pm \sqrt{4 \cos^2 \theta - 4}}{2} \]
\[ x = \cos \theta \pm \sqrt{\cos^2 \theta - 1} \]
Since \( \sqrt{\cos^2 \theta - 1} = i \sin \theta \), we have:
\[ x = \cos \theta \pm i \sin \theta \]
Thus, we can express \( x \) as:
\[ x = e^{i \theta} \quad \text{or} \quad x = e^{-i \theta} \]

### Step 3: Solving for \( y \)

Similarly, for the second equation:
\[ 2 \cos \phi = y + \frac{1}{y} \]
Following the same steps, we multiply by \( y \):
\[ 2 \cos \phi \cdot y = y^2 + 1 \]
Rearranging gives:
\[ y^2 - 2 \cos \phi \cdot y + 1 = 0 \]
Using the quadratic formula again:
\[ y = \frac{2 \cos \phi \pm \sqrt{(2 \cos \phi)^2 - 4}}{2} \]
This simplifies to:
\[ y = \cos \phi \pm i \sin \phi \]
Thus, we can express \( y \) as:
\[ y = e^{i \phi} \quad \text{or} \quad y = e^{-i \phi} \]

### Step 4: Finding \( \frac{x}{y} + \frac{y}{x} \)

Now we calculate:
\[ \frac{x}{y} + \frac{y}{x} \]
Assuming \( x = e^{i \theta} \) and \( y = e^{i \phi} \):
\[ \frac{x}{y} = \frac{e^{i \theta}}{e^{i \phi}} = e^{i(\theta - \phi)} \]
And:
\[ \frac{y}{x} = \frac{e^{i \phi}}{e^{i \theta}} = e^{i(\phi - \theta)} = e^{-i(\theta - \phi)} \]
Thus:
\[ \frac{x}{y} + \frac{y}{x} = e^{i(\theta - \phi)} + e^{-i(\theta - \phi)} = 2 \cos(\theta - \phi) \]

### Conclusion

Therefore, we have shown that:
\[ \frac{x}{y} + \frac{y}{x} = 2 \cos(\theta - \phi) \]