Solution:

Given equations are sin x sin y = sin (x-y) and |x| |y| = 1.

To find: Number of pairs (x, y) satisfying the given equations.

1. Simplifying the equations:

From the second equation, we get:

|x| = 1/|y|

If y > 0, then |x| = 1/y and x can take any value

If y < 0,="" then="" |x|="-1/y" and="" x="" can="" take="" values="" only="" in="" the="" range="" (-π,="" 0)="" and="" (0,="" />

From the first equation, we get:

sin x sin y = sin x cos y - cos x sin y

sin x sin y + cos x sin y = sin x cos y

sin (x+y) = sin x cos y / sin y

If sin y ≠ 0, then

sin (x+y) / sin y = sin x / cos y

tan y = cos y / tan (x+y)

tan y = cot x - cot y

If sin y = 0, then either y = kπ or x = kπ (k ∈ Z)

2. Finding the solutions:

Case 1: sin y ≠ 0

tan y = cot x - cot y

tan y + cot y = cot x

2 cot y = cot (x+y) + cot (x-y)

cot x + cot y = cot (x+y) cot (x-y)

Let t = cot x

Then cot y = (t±√(t²-4))/2

For each value of t, there are two possible values of cot y

Hence, there are 2n solutions in this case, where n is the number of values of t for which t² ≥ 4

Case 2: sin y = 0

y = kπ or x = kπ (k ∈ Z)

If y = kπ, then sin x sin y = 0 implies sin x = 0 or x = mπ (m ∈ Z)

If x = kπ, then sin x sin y = 0 implies sin y = 0 or y = mπ (m ∈ Z)

Hence, there are 6 solutions in this case

3. Final Answer:

Total number of solutions = 2n + 6

Since t² ≥ 4 for all real values of t, there are 2 possible values of t for each value of t.

Hence, n = 2 and the total number of solutions = 2n + 6 = 4 + 6 = 10.

Therefore, the correct option is (D) 6.