Question

tan y dx + sec² y tan x dy = 0

Answer 1

We have, 
\[\tan y dx + \sec^2 y \tan x dy = 0\]
\[ \Rightarrow \sec^2 y \tan x dy = - \tan y dx\]
\[ \Rightarrow \frac{\sec^2 y}{\tan y} dy = - \frac{1}{\tan x}dx\]
\[ \Rightarrow \frac{1}{\cos^2 y} \times \frac{\cos y}{\sin y}dy = - \cot x dx\]
\[ \Rightarrow \frac{1}{\sin y \cos y}dy = - \cot x dx\]
\[ \Rightarrow \frac{2}{\sin 2y}dy = - \cot x dx\]
\[ \Rightarrow 2 \text{ cosec } 2y dy = - \cot x dx\]
Integrating both sides, we get
\[2\int\text{ cosec }2y dy = - \int\cot x dx\]
\[ \Rightarrow \log \tan x = - \log \sin x = \log C\]
\[ \Rightarrow \log \tan x + \log \sin x = \log C\]
\[ \Rightarrow \log \left( \tan x \times \sin x \right) = \log C\]
\[ \Rightarrow \tan x \times \sin x = C\]