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Question:

$\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y=0$

Solution:

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.