Differential Equations — Class 12 Maths Solution

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

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

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

……(i)
On integrating both sides, we have
$\int {\frac{{{{\sec }^2}x}}{{\tan x}}} dx = - \int {\frac{{{{\sec }^2}y}}{{\tan y}}} dy$

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

we get
${\sec ^2}xdx = dt \Rightarrow {\sec ^2}xdx = dt$

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

we get
${\sec ^2}ydy = du$

On substituting these values in Eq. (i),

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

log $t = - \log u + \log k$
$\Rightarrow$ $\log (t \cdot u) = \log k$

$\Rightarrow$ $\log (\tan x\tan y) = \log k$

$\Rightarrow$ $\tan x\tan y = k$