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164 . The solution of the differential equation $\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y=0$ is
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Verified Answer
The correct answer is:
$\tan y \cdot \tan x=C$
Given, $\frac{\sec ^{2} x}{\tan x} d x=-\frac{\sec ^{2} y}{\tan y} d y$
$$
\Rightarrow \int \frac{\sec ^{2} x}{\tan x} d x=-\int \frac{\sec ^{2} y}{\tan y} d y
$$
Put $\quad \tan x=u \Rightarrow \sec ^{2} x d x=d u$
and $\quad \tan y=v \Rightarrow \sec ^{2} y d y=d v$
$\therefore \quad \int \frac{d u}{u}=-\int \frac{d v}{v}$
$\Rightarrow \log u=-\log v+\log C \Rightarrow w v=C \Rightarrow \tan x \cdot \tan y=C$
$$
\Rightarrow \int \frac{\sec ^{2} x}{\tan x} d x=-\int \frac{\sec ^{2} y}{\tan y} d y
$$
Put $\quad \tan x=u \Rightarrow \sec ^{2} x d x=d u$
and $\quad \tan y=v \Rightarrow \sec ^{2} y d y=d v$
$\therefore \quad \int \frac{d u}{u}=-\int \frac{d v}{v}$
$\Rightarrow \log u=-\log v+\log C \Rightarrow w v=C \Rightarrow \tan x \cdot \tan y=C$
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