Question

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

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

Answer 1

The correct option is B $\tan x\cdot \tan y=k$

Given:
$\tan y{\sec }^{2}xdx+\tan x{\sec }^{2}ydy=0\Rightarrow \frac{{\sec }^{2}x}{\tan x}dx+\frac{{\sec }^{2}y}{\tan y}dy=0\Rightarrow \frac{{\sec }^{2}x}{\tan x}dx=-\frac{{\sec }^{2}y}{\tan y}dy$
Integrating on both sides, we get
$\Rightarrow \int \frac{{\sec }^{2}x}{\tan x}dx=-\int \frac{{\sec }^{2}y}{\tan y}dy$
Taking $\tan x=u$, $\tan y=v$, then
$\Rightarrow {\sec }^{2}xdx=du,{\sec }^{2}ydy=dv$
Now $\Rightarrow \int \frac{1}{u}du=-\int \frac{1}{v}dv$
$\Rightarrow \ln |u|=-\ln |v|+c$
$\Rightarrow \ln |u|+\ln |v|=c$
$\Rightarrow \ln |uv|=c$
$\Rightarrow \ln |\tan x\cdot \tan y|=c$
$\Rightarrow |\tan x\cdot \tan y|=e^c$
$\Rightarrow \tan x\cdot \tan y=\pm e^c=k$