Let `x= sum_(n=0)^oo (-1)^n (tantheta)^(2n)` and ` y = sum_(n=0)^oo (costheta)^(2n)` qhere `theta in (0,pi/4)`, then

Options

A. (1)x(y+1)=1
B. (2)y(1-x) = 1
C. (3)y(x-1)=1
D. (4)y(1+x) = 1

Answer

To solve the given problem, we need to evaluate the sums \( x \) and \( y \) and then find a relationship between them. Let's go through the steps systematically.

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

The expression for \( x \) is given by:

\[ x = \sum_{n=0}^{\infty} (-1)^n (\tan \theta)^{2n} \]

This is a geometric series with the first term \( a = 1 \) and the common ratio \( r = -\tan^2 \theta \). The series converges when \( |r| < 1 \), which is true since \( \theta \in (0, \frac{\pi}{4}) \) implies \( \tan \theta < 1 \).

The sum of an infinite geometric series is given by:

\[ S = \frac{a}{1 - r} \]

Substituting our values:

\[ x = \frac{1}{1 - (-\tan^2 \theta)} = \frac{1}{1 + \tan^2 \theta} \]

Using the identity \( 1 + \tan^2 \theta = \sec^2 \theta \), we can simplify \( x \):

\[ x = \frac{1}{\sec^2 \theta} = \cos^2 \theta \]

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

The expression for \( y \) is given by:

\[ y = \sum_{n=0}^{\infty} (\cos \theta)^{2n} \]

This is also a geometric series with the first term \( a = 1 \) and the common ratio \( r = \cos^2 \theta \). The series converges since \( \cos \theta < 1 \) for \( \theta \in (0, \frac{\pi}{4}) \).

Using the formula for the sum of a geometric series:

\[ y = \frac{1}{1 - \cos^2 \theta} \]

Using the identity \( 1 - \cos^2 \theta = \sin^2 \theta \), we can simplify \( y \):

\[ y = \frac{1}{\sin^2 \theta} \]

### Step 3: Relate \( x \) and \( y \)

Now we have:

\[ x = \cos^2 \theta \quad \text{and} \quad y = \frac{1}{\sin^2 \theta} \]

Next, we will find a relationship between \( x \) and \( y \). We know:

\[ \sin^2 \theta + \cos^2 \theta = 1 \]

From the expressions for \( x \) and \( y \), we can write:

\[ 1 = \sin^2 \theta + \cos^2 \theta \]

Substituting \( \sin^2 \theta = \frac{1}{y} \) and \( \cos^2 \theta = x \):

\[ 1 = \frac{1}{y} + x \]

Rearranging gives us:

\[ x + \frac{1}{y} = 1 \]

Multiplying through by \( y \):

\[ xy + 1 = y \]

Rearranging gives us:

\[ y(1 - x) = 1 \]

### Final Result

Thus, the final relationship we derived is:

\[ y(1 - x) = 1 \]

### Conclusion

The correct option from the given choices is:

**Option 2: \( y(1 - x) = 1 \)**