`"For "0ltthetalt(pi)/(2)`, if
`x=sum_(n=0)^(oo)cos^(2n)theta,y=sum_(n=0)^(oo)sin^(2n)phi,z=sum_(n=0)^(oo)cos^(2n)thetasin^(2n)phi`, then

Options

A. xy=zx+zy+z
B. xy=zx+zy-z
C. xy+yz+zx=z
D. none of these

Explanation

To solve the problem, we need to evaluate the sums \( x \), \( y \), and \( z \) defined in the question. Let's break down the solution step by step.

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

The sum \( x \) is defined as:
\[ x = \sum_{n=0}^{\infty} \cos^{2n} \theta \]
This is a geometric series where the first term \( a = 1 \) and the common ratio \( r = \cos^2 \theta \).

Using the formula for the sum of an infinite geometric series:
\[ S = \frac{a}{1 - r} \]
we can write:
\[ x = \frac{1}{1 - \cos^2 \theta} \]
Since \( 1 - \cos^2 \theta = \sin^2 \theta \), we have:
\[ x = \frac{1}{\sin^2 \theta} \]

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

The sum \( y \) is defined as:
\[ y = \sum_{n=0}^{\infty} \sin^{2n} \phi \]
This is also a geometric series with first term \( a = 1 \) and common ratio \( r = \sin^2 \phi \).

Using the same formula for the sum:
\[ y = \frac{1}{1 - \sin^2 \phi} \]
Since \( 1 - \sin^2 \phi = \cos^2 \phi \), we have:
\[ y = \frac{1}{\cos^2 \phi} \]

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

The sum \( z \) is defined as:
\[ z = \sum_{n=0}^{\infty} \cos^{2n} \theta \sin^{2n} \phi \]
This can also be treated as a geometric series where the first term \( a = 1 \) and the common ratio \( r = \cos^2 \theta \sin^2 \phi \).

Using the formula for the sum:
\[ z = \frac{1}{1 - \cos^2 \theta \sin^2 \phi} \]

### Step 4: Relate \( x \), \( y \), and \( z \)

Now we have:
\[ x = \frac{1}{\sin^2 \theta}, \quad y = \frac{1}{\cos^2 \phi}, \quad z = \frac{1}{1 - \cos^2 \theta \sin^2 \phi} \]

We can express \( \cos^2 \theta \) and \( \sin^2 \phi \) in terms of \( x \) and \( y \):
\[ \cos^2 \theta = \frac{1}{x}, \quad \sin^2 \phi = \frac{1}{y} \]

Substituting these into the expression for \( z \):
\[ z = \frac{1}{1 - \left(\frac{1}{x}\right)\left(\frac{1}{y}\right)} = \frac{1}{1 - \frac{1}{xy}} = \frac{xy}{xy - 1} \]

### Step 5: Establish the relationship

From the above, we can establish the relationship:
\[ x + y - 1 = xy \]

### Conclusion

Thus, the final relationship we derived is:
\[ xy = x + y - 1 \]