Question

The solution of the differential equation ${\sec }^{2}x\tan ydx+{\sec }^{2}y\tan xdy=0$ is

• A. $\tan y\tan x=c$
• B. $\frac{\tan y}{\tan x}=c$
• C. $\frac{{\tan }^{2}x}{\tan y}=c$
• D. None of these

Answer 1

Answer: (A)

$\tan y\tan x=c$

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

On separating the variables (dividing the equation by \tan x \tan y)

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

On integrating both sides, we get

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

Put $\tan x=u\Rightarrow {\sec }^{2}x.dx=du$ and $\tan y=v\Rightarrow {\sec }^{2}y.dy=dv$

$\therefore \int \frac{du}{u}=-\int \frac{dv}{v}$

$\Rightarrow logu=-logv+logc$

$\Rightarrow u=\frac{c}{v}\Rightarrow u.v=c$

$\therefore \tan x.\tan y=c$