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. x(y+1)=1
B. y(1-x) = 1
C. y(x-1)=1
D. y(1+x) = 1

Answer

To solve the problem, we need to evaluate the sums \( x \) and \( y \) given by:

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

where \( \theta \in (0, \frac{\pi}{4}) \).

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

The series for \( x \) is an alternating geometric series. The general form of an alternating geometric series is:

\[ \sum_{n=0}^{\infty} (-1)^n a^n = \frac{1}{1 + a} \quad \text{for } |a| < 1 \]

In our case, \( a = \tan^2 \theta \). Since \( \theta \in (0, \frac{\pi}{4}) \), we have \( \tan \theta < 1 \), thus \( \tan^2 \theta < 1 \).

Using the formula for the sum of the series, we get:

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

### Step 2: Simplify \( x \)

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

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

### Step 3: Evaluate \( y \)

The series for \( y \) is a standard geometric series:

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

The sum of a geometric series is given by:

\[ \sum_{n=0}^{\infty} r^n = \frac{1}{1 - r} \quad \text{for } |r| < 1 \]

Here, \( r = \cos^2 \theta \), and since \( \theta \in (0, \frac{\pi}{4}) \), we have \( \cos \theta < 1 \) and thus \( \cos^2 \theta < 1 \). Therefore:

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

### Step 4: Simplify \( y \)

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

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

### Step 5: Find \( x + \frac{1}{y} \)

Now we need to compute \( x + \frac{1}{y} \):

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

Thus,

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

Using the Pythagorean identity:

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

### Conclusion

We have found that:

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

This leads us to the final result:

\[ xy + 1 = y \implies y(1 - x) = 1 \]

Thus, the answer is:

\[ \boxed{1} \]