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The general solution of the differential equation $\tan x \tan y d x+\cos ^2 x \operatorname{cosec}^2 y d y=0$ is
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Verified Answer
The correct answer is:
$\tan ^2 x-\cot ^2 y=C$
Given differential equation
$\begin{aligned}
& \tan x \cdot \tan y d x+\cos ^2 x \cdot \operatorname{cosec}^2 y d y=0 \\
\Rightarrow & \frac{\tan x}{\cos ^2 x} d x+\frac{\operatorname{cosec}^2 y}{\tan y} d y=0 \\
\Rightarrow & \int \tan x \cdot \sec ^2 x d x+\int \cot y \cdot \operatorname{cosec}^2 y d y=0 \\
\Rightarrow & \tan ^2 x-\cot ^2 y=C .
\end{aligned}$
$\begin{aligned}
& \tan x \cdot \tan y d x+\cos ^2 x \cdot \operatorname{cosec}^2 y d y=0 \\
\Rightarrow & \frac{\tan x}{\cos ^2 x} d x+\frac{\operatorname{cosec}^2 y}{\tan y} d y=0 \\
\Rightarrow & \int \tan x \cdot \sec ^2 x d x+\int \cot y \cdot \operatorname{cosec}^2 y d y=0 \\
\Rightarrow & \tan ^2 x-\cot ^2 y=C .
\end{aligned}$
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