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Solve the following differential equation
$\sec ^2 x \tan y d x+\sec ^2 y \tan x d y=0$
Solution: $\sec ^2 x \tan y d x+\sec ^2 y \tan x d y=0$
$\therefore \frac{\sec ^2 x}{\tan x} d x+\square=0$
Integrating, we get
$\square+\int \frac{\sec ^2 y}{\tan y} d y=\log c$
Each of these integral is of the type
$\int \frac{ f ^{\prime}(x)}{ f (x)} d x=\log | f ( x )|+\log C$
$\therefore$ the general solution is
$\square+\log |\tan y|=\log c$
$\therefore \log |\tan x \cdot \tan y|=\log c$
$\square$
This is the general solution.