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Solve $\sec^2 x \cdot \tan y \, dx + \sec^2 y \cdot \tan x \, dy = 0$:
A$\tan x \cdot \tan y = c$
B$\tan x + \tan y = c$
C$\sec x + \sec y = c$
D$\sin x \cdot \sin y = c$
Answer & Solution
Correct answer: A. $\tan x \cdot \tan y = c$
Rewrite: $(\sec^2 y/\tan y)dy + (\sec^2 x/\tan x)dx = 0$. Recognise each as a $du/u$: integrating gives $\log\tan y + \log\tan x = \log c$ ⇒ $\tan x \cdot \tan y = c$.
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