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We are given the equation:

\[
\frac{\cos(x - y)}{\cos(x + y)} = \frac{-9}{7}
\]

We are asked to find the value of \(\tan x \cdot \tan y\).

### Step 1: Use cosine sum and difference identities

We apply the cosine angle difference and sum formulas:
- \(\cos(x - y) = \cos x \cos y + \sin x \sin y\)
- \(\cos(x + y) = \cos x \cos y - \sin x \sin y\)

So, the equation becomes:

\[
\frac{\cos x \cos y + \sin x \sin y}{\cos x \cos y - \sin x \sin y} = \frac{-9}{7}
\]

### Step 2: Simplify

Let’s denote \(\cos x \cos y = A\) and \(\sin x \sin y = B\) for simplicity. Thus, the equation becomes:

\[
\frac{A + B}{A - B} = \frac{-9}{7}
\]

Cross-multiply:

\[
7(A + B) = -9(A - B)
\]

Expanding both sides:

\[
7A + 7B = -9A + 9B
\]

### Step 3: Solve for A and B

Now, combine like terms:

\[
7A + 9A = 9B - 7B
\]

\[
16A = 2B
\]

\[
B = 8A
\]

Since \(A = \cos x \cos y\) and \(B = \sin x \sin y\), we have:

\[
\sin x \sin y = 8 \cos x \cos y
\]

### Step 4: Find \(\tan x \cdot \tan y\)

Using the identity \(\tan x \cdot \tan y = \frac{\sin x \sin y}{\cos x \cos y}\), we can substitute:

\[
\tan x \cdot \tan y = \frac{8 \cos x \cos y}{\cos x \cos y} = 8
\]

Thus, the value of \(\tan x \cdot \tan y\) is:

\[
\boxed{8}
\]

Would you like more details on any step?

### Related questions:
1. How are the sum and difference identities for cosine derived?
2. What happens if the given ratio of \(\frac{-9}{7}\) was different?
3. How do you calculate \(\tan(x+y)\) given \(\tan x\) and \(\tan y\)?
4. Can we solve this using a different trigonometric identity?
5. How would this problem change if sine identities were used instead of cosine?

### Tip: 
When dealing with trigonometric identities, it's helpful to apply standard formulas for sums, differences, and products to simplify complex ratios.

Given that cos(x - y) / cos(x + y) = -9 / 7. Find the value of tan x * tan y.

Learn how to find the value of tan x * tan y given cos(x - y) / cos(x + y) = -9/7. This problem uses trigonometric identities such as cosine sum and difference formulas, suitable for students in Grades 10-12.

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