Given that `0 lt x lt (pi)/(4) and (pi)/(4) lt y lt (pi)/(2) and sum_(k = 0)^(oo) (-1)^(k) tan^(2k) x = p, sum_(k = 0)^(oo) (-1)^(k) cot^(2k) y = q`, then `sum_(k = 0)^(oo) tan^(2k) x cot^(2k) y` is

Options

A. `(1)/(p) + (1)/(q) - (1)/(pq)`
B. `(1)/({(1)/(p) + (1)/(q) - (1)/(pq)})`
C. p + q - pq
D. p + q + pq

Answer

To solve the problem, we need to evaluate the series \( S = \sum_{k=0}^{\infty} \tan^{2k} x \cot^{2k} y \).

### Step 1: Express the series in terms of \( \tan^2 x \) and \( \cot^2 y \)

We can rewrite the series as:
\[ S = \sum_{k=0}^{\infty} (\tan^2 x \cot^2 y)^k \]

### Step 2: Identify the common ratio

The common ratio of the series is \( r = \tan^2 x \cot^2 y \).

### Step 3: Check the convergence condition

Since \( 0 < x < \frac{\pi}{4} \) implies \( \tan^2 x < 1 \) and \( \frac{\pi}{4} < y < \frac{\pi}{2} \) implies \( \cot^2 y < 1 \), we have:
\[ \tan^2 x \cot^2 y < 1 \]
Thus, the series converges.

### Step 4: Use the formula for the sum of an infinite geometric series

The sum of an infinite geometric series is given by:
\[ S = \frac{1}{1 - r} \]
where \( r \) is the common ratio. Therefore, we have:
\[ S = \frac{1}{1 - \tan^2 x \cot^2 y} \]

### Step 5: Substitute \( \tan^2 x \) and \( \cot^2 y \)

From the given information, we know:
\[ p = \sum_{k=0}^{\infty} (-1)^k \tan^{2k} x = \frac{1}{1 + \tan^2 x} \]
and
\[ q = \sum_{k=0}^{\infty} (-1)^k \cot^{2k} y = \frac{1}{1 + \cot^2 y} \]

### Step 6: Express \( \tan^2 x \) and \( \cot^2 y \) in terms of \( p \) and \( q \)

From the expressions for \( p \) and \( q \):
\[ \tan^2 x = \frac{1}{p} - 1 \]
\[ \cot^2 y = \frac{1}{q} - 1 \]

### Step 7: Substitute back into the expression for \( S \)

Now substituting these into the expression for \( S \):
\[ S = \frac{1}{1 - \left(\frac{1}{p} - 1\right)\left(\frac{1}{q} - 1\right)} \]

### Step 8: Simplify the expression

Calculating the product:
\[ \left(\frac{1}{p} - 1\right)\left(\frac{1}{q} - 1\right) = \frac{1}{pq} - \left(\frac{1}{p} + \frac{1}{q}\right) + 1 \]
Thus,
\[ S = \frac{1}{1 - \left(\frac{1}{pq} - \left(\frac{1}{p} + \frac{1}{q}\right) + 1\right)} \]
This simplifies to:
\[ S = \frac{pq}{p + q - 1} \]

### Final Result

Thus, the final result is:
\[ \sum_{k=0}^{\infty} \tan^{2k} x \cot^{2k} y = \frac{pq}{p + q - 1} \]