Question

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

Answer 1

Consider the following Equation.

$se{c}^{2}xtanydy+{\sec }^{2}ytanxdx=0$

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

$u=\frac{1}{{\sec }^{2}x}$

$dx=\frac{{\sec }^{2}x}{2\tan x}du$

Similarly,

$v=\frac{1}{{\sec }^{2}y}$

$dy=\frac{{\sec }^{2}y}{2\tan y}dv$

Therefore,

$\underset{}{\overset{}{\int }}\frac{\tan x}{{\sec }^{2}x}\frac{{\sec }^{2}x}{2\tan x}du=\underset{}{\overset{}{\int }}\frac{\tan y}{{\sec }^{2}y}\frac{{\sec }^{2}y}{2\tan y}dv$

$\frac{1}{2}\underset{}{\overset{}{\int }}1du=-\frac{1}{2}\underset{}{\overset{}{\int }}1dv$

$\frac{u}{2}=-\frac{v}{2}+C$

$\frac{1}{{2\sec }^{2}x}=-\frac{1}{{2\sec }^{2}y}+C$

${\cos }^{2}x+{\cos }^{2}y=2C$

Hence, this is the correct answer.