Question

${\sec }^{2}x\cdot \tan ydx+{\sec }^{2}y\cdot \tan xdy=0$

Answer 1

${\sec }^{2}x.\tan ydx\ne {\sec }^{2}y.\tan xdy=0$

Divide both side by $\tan y.\tan x$

$\begin{array}{c}\frac{{\sec }^{2}x\tan ydx+{\sec }^{2}y.\tan xdy}{\tan y.\tan x}=\frac{0}{\tan x.\tan y}\\ \frac{{\sec }^{2}x\tan ydx}{\tan y\tan x}+\frac{{\sec }^{2}y.\tan xdy}{\tan y.\tan x}=0\\ \frac{{\sec }^{2}x}{\tan x}dx+\frac{{\sec }^{2}y}{\tan y}dy=0\end{array}$

Integrating both side

$\begin{array}{ccc}\int \frac{{\sec }^{2}x}{\tan x}dx+\int \frac{{\sec }^{2}y}{\tan y}dy=0& put& \\ \int \frac{1}{u}du+\int \frac{1}{v}dv=0& \begin{array}{c}\tan x=4\\ {\sec }^{2}xdx=dy\end{array}& \\ \log \left|u\right|+\log \left|v\right|=\log c& \begin{array}{c}\tan y=v\\ {\sec }^{2}ydy=dv\end{array}& \\ \log \left|\tan x\right|+\log \left|\tan y\right|=\log c& usingformulae& \\ \log \left|\tan x.\tan y\right|=\log c& & \log m+\log n=\log mn\\ \tan x.\tan y=c& & \end{array}$