Solve `sec^(2)x tany dx+sec^(2)y tanx dy=0`

Explanation

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

### Step 1: Rearranging the Equation
We start with the given equation:
\[ \sec^2 x \tan y \, dx + \sec^2 y \tan x \, dy = 0 \]
We can rearrange this to isolate \( dx \) and \( dy \):
\[ \sec^2 x \tan y \, dx = -\sec^2 y \tan x \, dy \]
This can be rewritten as:
\[ \frac{dx}{dy} = -\frac{\sec^2 y \tan x}{\sec^2 x \tan y} \]

### Step 2: Separating Variables
We can separate the variables \( x \) and \( y \):
\[ \frac{\sec^2 x \tan y}{\tan x} \, dx + \frac{\sec^2 y}{\tan y} \, dy = 0 \]
This means we can express it as:
\[ \frac{\sec^2 x}{\tan x} \, dx + \frac{\sec^2 y}{\tan y} \, dy = 0 \]

### Step 3: Integrating Both Sides
Now, we integrate both sides:
\[ \int \frac{\sec^2 x}{\tan x} \, dx + \int \frac{\sec^2 y}{\tan y} \, dy = 0 \]
The integral of \( \frac{\sec^2 x}{\tan x} \) is \( \log |\tan x| \), and similarly for \( y \):
\[ \log |\tan x| + \log |\tan y| = C \]
where \( C \) is a constant of integration.

### Step 4: Simplifying the Result
Using the properties of logarithms, we can combine the logs:
\[ \log |\tan x \tan y| = C \]
Exponentiating both sides gives:
\[ |\tan x \tan y| = e^C \]
Let \( k = e^C \), so we have:
\[ \tan x \tan y = k \]

### Final Answer
Thus, the solution to the differential equation is:
\[ \tan x \tan y = C \]
where \( C \) is a constant.

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