Question

Solve ${\sec }^{2}x\tan ydy+{\sec }^{2}y\tan xdx=0$

Answer 1

Simplifying ${\sec }^{2}x\tan ydy+{\sec }^{2}y\tan xdx=0$

${\sec }^{2}x\tan ydy=-{\sec }^{2}y\tan xdx$

$\frac{\tan y}{{\sec }^{2}y}dy=-\frac{\tan x}{{\sec }^{2}x}dx$

$\frac{\tan y}{{\sec }^{2}y}dy=-\frac{\tan x}{{\sec }^{2}x}dx$

$\frac{\frac{\sin y}{\cos y}}{\frac{1}{{\cos }^{2}y}}dy=-\frac{\frac{\sin x}{\cos x}}{\frac{1}{{\cos }^{2}x}}dx$

$\cos y\sin ydy=-\sin x\cos xdx$

$\int \cos y\sin ydy=-\int \sin x\cos xdx$

Put$t=\sin y,dt=\cos ydy$

And, $u=\cos x,du=-\sin xdx$

$\int tdt=\int udu$

$\frac{t^2}{2}=\frac{u^2}{2}+c$

$\frac{{\sin }^{2}y}{2}=\frac{{\cos }^{2}x}{2}$

${\sin }^{2}y-{\cos }^{2}x=0$