The solution of different equation `sec^(2)x tanydx+ sec^(2)y tan xdy= 0 AA x, y in R- (2n+1) (pi)/(2), n in I` is

Options

A. `tan x= tan y`
B. `tan x tan y=c, AA x, y in R`
C. `tan x tan y=c`
D. None of these

Answer

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 differential equation:
\[ \sec^2 x \tan y \, dx + \sec^2 y \tan x \, dy = 0 \]
Rearranging gives:
\[ \sec^2 x \tan y \, dx = -\sec^2 y \tan x \, dy \]

### Step 2: Separating Variables
We can separate the variables \(x\) and \(y\) by dividing both sides by \(\sec^2 y \tan x \tan y\):
\[ \frac{\sec^2 x}{\tan x} \, dx = -\frac{\sec^2 y}{\tan y} \, dy \]
This can be rewritten as:
\[ \frac{\sec^2 x}{\tan x} \, dx = -\frac{\sec^2 y}{\tan y} \, dy \]

### Step 3: Integrating Both Sides
Now we integrate both sides:
\[ \int \frac{\sec^2 x}{\tan x} \, dx = -\int \frac{\sec^2 y}{\tan y} \, dy \]
Using the substitution \( u = \tan x \) and \( v = \tan y \), we have:
\[ \int \frac{1}{u} \, du = -\int \frac{1}{v} \, dv \]
This results in:
\[ \ln |u| = -\ln |v| + C \]

### Step 4: Simplifying the Result
Exponentiating both sides gives:
\[ |u| \cdot |v| = e^C \]
Let \( k = e^C \) (a constant), we can write:
\[ \tan x \cdot \tan y = k \]

### Step 5: Final Solution
Thus, the general solution to the differential equation is:
\[ \tan x \cdot \tan y = C \]
where \( C \) is a constant.