Find the general solution of the differential equations `sec^2xtany dx+sec^2ytanx 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 by rearranging the given equation:
\[ \sec^2 x \tan y \, dx = -\sec^2 y \tan x \, dy \]

### Step 2: Separating Variables
Next, we separate the variables:
\[ \frac{\sec^2 x}{\tan x} \, dx = -\frac{\sec^2 y}{\tan y} \, dy \]
This can be rewritten as:
\[ \frac{\sec^2 x}{\tan x} \, dx = -\frac{\sec^2 y}{\tan y} \, dy \]

### Step 3: Substitutions
Now, we will make substitutions for easier integration. Let:
\[ \tan x = t \quad \text{and} \quad \tan y = m \]
Then, we differentiate both sides:
\[ \sec^2 x \, dx = dt \quad \text{and} \quad \sec^2 y \, dy = dm \]
Substituting these into the separated equation gives:
\[ \frac{dt}{t} = -\frac{dm}{m} \]

### Step 4: Integrating Both Sides
Now we integrate both sides:
\[ \int \frac{dt}{t} = -\int \frac{dm}{m} \]
This results in:
\[ \ln |t| = -\ln |m| + C \]
where \( C \) is the constant of integration.

### Step 5: Simplifying the Result
Using properties of logarithms, we can rewrite this as:
\[ \ln |t| + \ln |m| = C \]
This simplifies to:
\[ \ln |tm| = C \]
Exponentiating both sides gives:
\[ tm = e^C \]
Letting \( k = e^C \), we have:
\[ \tan x \tan y = k \]

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

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