The solution of differential equation $\tan y \sec^2 x dx + \tan x \sec^2 y dy = 0$ is

$\tan x \cdot \tan y = k$

The correct answer is Option (4) → $\tan x \cdot \tan y = k$ ##

Given that, $\tan y \sec^2 x dx + \tan x \sec^2 y dy = 0$

$⇒\tan y \sec^2 x dx = -\tan x \sec^2 y dy$

$⇒ \frac{\sec^2 x}{\tan x} dx = \frac{-\sec^2 y}{\tan y} dy \quad \dots(i)$

On integrating both sides, we get

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

Put $\tan x = t$ in LHS integral, we get

$\sec^2 x \, dx = dt$

and $\tan y = u$ in RHS integral, we get

$\sec^2 y \, dy = du$

On substituting these values in Eq. (i), we get

$\int \frac{dt}{t} = -\int \frac{du}{u}$

$⇒ \log t = -\log u + \log k$

$⇒ \log t + \log u = \log k$

$⇒ \log(t \cdot u) = \log k \quad [∵\log x + \log y = \log (xy)]$

$⇒ \log(\tan x \tan y) = \log k$

$⇒ \tan x \tan y = k$