Question

Solution of the differential equation
$\tan y.{\sec }^{2}xdx+\tan x.{\sec }^{2}ydy=0$ is

• A. $\tan x+\tan y=k$
• B. $\tan x-\tan y=k$
• C. $\frac{\tan x}{\tan y}=k$
• D. $\tan x.\tan y=k$

Answer 1

Answer: (D)

$\tan x.\tan y=k$

$\tan y.{\sec }^{2}xdx+\tan x.{\sec }^{2}ydy=0$
$\Rightarrow \tan y.{\sec }^{2}xdx=-\tan x.{\sec }^{2}ydy$
$\Rightarrow \frac{{\sec }^{2}xdx}{\tan x}=-\frac{{\sec }^{2}ydx}{\tan y}$
Integrating, $\int \frac{{\sec }^{2}xdx}{\tan x}=-\int \frac{{\sec }^{2}ydx}{\tan y}$
$\Rightarrow \int \frac{d\left(\tan x\right)}{\tan x}=-\int \frac{d\left(\tan y\right)}{\tan y}$
$\Rightarrow \ln \left|\tan x\right|=-\ln \left|\tan x\right|+c$
$\Rightarrow \ln \left|\tan x\right|+\ln \left|\tan x\right|=c$
$\Rightarrow \ln |\tan x.\tan y|=c$
$\Rightarrow \tan x\tan y=e^c$
$\Rightarrow \tan x\tan y=k$ (where $e^c=k$)