$\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y=0$
The given differential equation is:
$\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y=0$
$\Rightarrow \frac{\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y}{\tan x \tan y}=0$
$\Rightarrow \frac{\sec ^{2} x}{\tan x} d x+\frac{\sec ^{2} y}{\tan y} d y=0$
$\Rightarrow \frac{\sec ^{2} x}{\tan x} d x=-\frac{\sec ^{2} y}{\tan y} d y$
Integrating both sides of this equation, we get:
$\int \frac{\sec ^{2} x}{\tan x} d x=-\int \frac{\sec ^{2} y}{\tan y} d y$ ...(1)
Let $\tan x=t$.
$\therefore \frac{d}{d x}(\tan x)=\frac{d t}{d x}$
$\Rightarrow \sec ^{2} x=\frac{d t}{d x}$
$\Rightarrow \sec ^{2} x d x=d t$
Now, $\int \frac{\sec ^{2} x}{\tan x} d x=\int \frac{1}{t} d t$
$=\log t$
$=\log (\tan x)$
Similarly, $\int \frac{\sec ^{2} x}{\tan x} d y=\log (\tan y)$
Substituting these values in equation (1), we get:
$\log (\tan x)=-\log (\tan y)+\log \mathrm{C}$
$\Rightarrow \log (\tan x)=\log \left(\frac{\mathrm{C}}{\tan y}\right)$
$\Rightarrow \tan x=\frac{\mathrm{C}}{\tan y}$
$\Rightarrow \tan x \tan y=\mathrm{C}$
This is the required general solution of the given differential equation.
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