If `0 lt theta, phi lt (pi)/(2), x = sum_(n=0)^(oo)cos^(2n)theta, y=sum_(n=0)^(oo)sin^(2n)phi` and `z=sum_(n=0)^(oo)cos^(2n)theta*sin^(2n)phi` then :

Options

A. `xy-z=(x+y)z`
B. `xy+yz+zx=z`
C. `xyz=4`
D. `xy+z=(x+y)z`

Answer

To solve the problem, we need to evaluate the sums \( x \), \( y \), and \( z \) defined as follows:

1. \( x = \sum_{n=0}^{\infty} \cos^{2n} \theta \)
2. \( y = \sum_{n=0}^{\infty} \sin^{2n} \phi \)
3. \( z = \sum_{n=0}^{\infty} \cos^{2n} \theta \sin^{2n} \phi \)

### Step 1: Calculate \( x \)

The series for \( x \) is a geometric series with the first term \( a = 1 \) and common ratio \( r = \cos^2 \theta \).

The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]

Thus, we have:
\[ x = \frac{1}{1 - \cos^2 \theta} \]

Using the identity \( 1 - \cos^2 \theta = \sin^2 \theta \), we can rewrite \( x \) as:
\[ x = \frac{1}{\sin^2 \theta} \]

### Step 2: Calculate \( y \)

Similarly, for \( y \), we have:
\[ y = \sum_{n=0}^{\infty} \sin^{2n} \phi \]

This is also a geometric series with first term \( a = 1 \) and common ratio \( r = \sin^2 \phi \). Thus:
\[ y = \frac{1}{1 - \sin^2 \phi} \]

Using the identity \( 1 - \sin^2 \phi = \cos^2 \phi \), we can rewrite \( y \) as:
\[ y = \frac{1}{\cos^2 \phi} \]

### Step 3: Calculate \( z \)

Now, for \( z \):
\[ z = \sum_{n=0}^{\infty} \cos^{2n} \theta \sin^{2n} \phi \]

This is also a geometric series with first term \( a = 1 \) and common ratio \( r = \cos^2 \theta \sin^2 \phi \). Thus:
\[ z = \frac{1}{1 - \cos^2 \theta \sin^2 \phi} \]

### Step 4: Substitute the values of \( x \) and \( y \)

Now substituting the expressions we found for \( x \) and \( y \):
\[ z = \frac{1}{1 - \left( \frac{1}{\sin^2 \theta} \right) \left( \frac{1}{\cos^2 \phi} \right)} \]

This simplifies to:
\[ z = \frac{1}{1 - \frac{1}{\sin^2 \theta \cos^2 \phi}} = \frac{\sin^2 \theta \cos^2 \phi}{\sin^2 \theta \cos^2 \phi - 1} \]

### Step 5: Final relationship

Now, we can express \( z \) in terms of \( x \) and \( y \):
\[ z = \frac{1}{\frac{1}{y} + \frac{1}{x} - \frac{1}{xy}} = \frac{xy}{x + y - 1} \]

Thus, we have the relationship:
\[ z + xy = x + y \]

### Conclusion

The final result is:
\[ z + xy = x + y \]