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 can follow these steps:

### Step 1: Rearranging the Equation
We can rearrange the given equation to separate the variables:
\[ \sec^2 x \tan y \, dx = -\sec^2 y \tan x \, dy \]
This can be rewritten as:
\[ \frac{dx}{\sec^2 y \tan x} = -\frac{dy}{\sec^2 x \tan y} \]

### Step 2: Integrating Both Sides
Now, we can integrate both sides. We will integrate the left side with respect to \( x \) and the right side with respect to \( y \):
\[ \int \frac{dx}{\sec^2 y \tan x} = -\int \frac{dy}{\sec^2 x \tan y} \]

### Step 3: Simplifying the Integrals
The left integral can be simplified:
\[ \int \frac{dx}{\sec^2 y \tan x} = \frac{1}{\sec^2 y} \int \frac{dx}{\tan x} = \frac{1}{\sec^2 y} \ln |\sin x| + C_1 \]
Similarly, the right integral can be simplified:
\[ -\int \frac{dy}{\sec^2 x \tan y} = -\frac{1}{\sec^2 x} \int \frac{dy}{\tan y} = -\frac{1}{\sec^2 x} \ln |\sin y| + C_2 \]

### Step 4: Combining the Results
Combining the results from both integrals, we have:
\[ \frac{1}{\sec^2 y} \ln |\sin x| + \frac{1}{\sec^2 x} \ln |\sin y| = C \]
where \( C \) is a constant.

### Step 5: Final Form
To express the solution in a more standard form, we can multiply through by \( \sec^2 x \sec^2 y \):
\[ \ln |\sin x| \sec^2 y + \ln |\sin y| \sec^2 x = C \sec^2 x \sec^2 y \]

This represents the general solution of the differential equation.