Solve each of the following differential equations
`tan x tan y dx + sec^2x sec^2 y dy=0`.

Answer

To solve the differential equation \( \tan x \tan y \, dx + \sec^2 x \sec^2 y \, dy = 0 \), we can follow these steps:

### Step 1: Rearrange the equation
We start by rearranging the equation to separate the variables \(x\) and \(y\):

\[ \tan x \tan y \, dx + \sec^2 x \sec^2 y \, dy = 0 \]

This can be rewritten as:

\[ \sec^2 x \sec^2 y \, dy = -\tan x \tan y \, dx \]

Now, we can separate the variables:

\[ \frac{dy}{\tan y} = -\frac{\tan x}{\sec^2 x} \, dx \]

### Step 2: Simplify the equation
We know that \( \sec^2 x = 1 + \tan^2 x \), and thus we can rewrite the right-hand side:

\[ \frac{dy}{\tan y} = -\tan x \cos^2 x \, dx \]

### Step 3: Integrate both sides
Now we can integrate both sides. The left side can be integrated as follows:

\[ \int \frac{dy}{\tan y} = \int \frac{1}{\sin y} \, dy = \ln |\sin y| + C_1 \]

For the right side, we have:

\[ \int -\tan x \cos^2 x \, dx \]

Using the identity \( \tan x = \frac{\sin x}{\cos x} \):

\[ \int -\frac{\sin x}{\cos x} \cos^2 x \, dx = -\int \sin x \cos x \, dx \]

This can be integrated as:

\[ -\frac{1}{2} \sin^2 x + C_2 \]

### Step 4: Combine the results
Now we can combine the results of the integrals:

\[ \ln |\sin y| = -\frac{1}{2} \sin^2 x + C \]

Where \(C = C_2 - C_1\).

### Step 5: Solve for \(y\)
To express \(y\) in terms of \(x\), we exponentiate both sides:

\[ |\sin y| = e^{-\frac{1}{2} \sin^2 x + C} = e^C e^{-\frac{1}{2} \sin^2 x} \]

Let \(k = e^C\), then:

\[ \sin y = k e^{-\frac{1}{2} \sin^2 x} \]

### Final Solution
Thus, the solution to the differential equation is:

\[ \sin y = k e^{-\frac{1}{2} \sin^2 x} \]