Question

tan y dx + sec² y tan x dy = 0

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \tan ydx+{\sec }^{2}y\tan xdy=0 \\ \Rightarrow {\sec }^{2}y\tan xdy=-\tan ydx \\ \Rightarrow \frac{{\sec }^{2}y}{\tan y}dy=-\frac{1}{\tan x}dx \\ \Rightarrow \frac{1}{{\cos }^{2}y}\times \frac{\cos y}{\sin y}dy=-\cot xdx \\ \Rightarrow \frac{1}{\sin y\cos y}dy=-\cot xdx \\ \Rightarrow \frac{2}{\sin 2y}dy=-\cot xdx \\ \Rightarrow 2\mathrm{cosec}2\mathrm{y}dy=-\cot xdx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ 2\int \mathrm{cosec}2\mathrm{y}\mathrm{d}\mathrm{y}=-\int \cot \mathrm{x}\mathrm{d}\mathrm{x} \\ \Rightarrow \log \tan x=-\log \sin x=\log C \\ \Rightarrow \log \tan x+\log \sin x=\log C \\ \Rightarrow \log \left(\tan x\times \sin x\right)=\log C \\ \Rightarrow \tan x\times \sin x=C \end{gathered}$$