If `0 lt theta lt pi/2, x= sum_(n=0)^(oo) cos^(2n) theta, y= sum_(n=0)^(oo) sin^(2n) theta` and `z=sum_(n=0)^(oo) cos^(2n) theta* Sin^(2n) theta`, then show `xyz=xy+z`.
(a) xyz = xz + y
(b) xyz = xy + z
(c) xyz = x + y + z
(d) xyz = yz + x

Answer

To solve the problem, we need to find the values of \( x \), \( y \), and \( z \) based on the given infinite series and then show that \( xyz = xy + z \).

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

Given:
\[ x = \sum_{n=0}^{\infty} \cos^{2n} \theta \]

This is a geometric series with the first term \( a = 1 \) and the common ratio \( r = \cos^2 \theta \). The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
Thus, we have:
\[ x = \frac{1}{1 - \cos^2 \theta} = \frac{1}{\sin^2 \theta} \]

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

Given:
\[ y = \sum_{n=0}^{\infty} \sin^{2n} \theta \]

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

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

Given:
\[ z = \sum_{n=0}^{\infty} \cos^{2n} \theta \sin^{2n} \theta \]

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

### Step 4: Show that \( xyz = xy + z \)

Now substituting the values of \( x \), \( y \), and \( z \):
\[ xyz = \left(\frac{1}{\sin^2 \theta}\right) \left(\frac{1}{\cos^2 \theta}\right) \left(\frac{1}{1 - \cos^2 \theta \sin^2 \theta}\right) \]
\[ = \frac{1}{\sin^2 \theta \cos^2 \theta (1 - \cos^2 \theta \sin^2 \theta)} \]

Now for \( xy + z \):
\[ xy = \left(\frac{1}{\sin^2 \theta}\right) \left(\frac{1}{\cos^2 \theta}\right) = \frac{1}{\sin^2 \theta \cos^2 \theta} \]
\[ xy + z = \frac{1}{\sin^2 \theta \cos^2 \theta} + \frac{1}{1 - \cos^2 \theta \sin^2 \theta} \]

Finding a common denominator for \( xy + z \):
\[ = \frac{1(1 - \cos^2 \theta \sin^2 \theta) + \sin^2 \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta (1 - \cos^2 \theta \sin^2 \theta)} \]
\[ = \frac{1 - \cos^2 \theta \sin^2 \theta + \sin^2 \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta (1 - \cos^2 \theta \sin^2 \theta)} \]
\[ = \frac{1}{\sin^2 \theta \cos^2 \theta (1 - \cos^2 \theta \sin^2 \theta)} \]

Thus, we have:
\[ xyz = xy + z \]

### Conclusion

We have shown that:
\[ xyz = xy + z \]
Therefore, the answer is:
**(b) \( xyz = xy + z \)**