Question

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

• A. $\tan x.\cot y=C$
• B. $\cot x.\tan y=C$
• C. $\tan x.\tan y=C$
• D. $\sin x.\cos y=C$

Answer 1

Answer: (C)

Given, ${\sec }^{2}y\cdot \tan ydx+{\sec }^{2}y\cdot \tan xdy=0$
On separating the variables, we get
$\Rightarrow {\sec }^{2}x\cdot \tan ydx=-{\sec }^{2}y\cdot \tan xdy$
$\Rightarrow \frac{{\sec }^{2}x}{\tan x}dx=-\frac{{\sec }^{2}y}{\tan y}dy$
On integrating both the sides, we get
$\int \frac{{\sec }^{2}x}{\tan x}dx=-\int \frac{{\sec }^{2}x}{\tan y}dy$
Let $\tan x=u$
$\Rightarrow {\sec }^{2}x=\frac{du}{dx}$
$\Rightarrow dx=\frac{du}{{\sec }^{2}y}$ and $\tan y=v$
$\Rightarrow {\sec }^{2}y=\frac{dv}{dy}$
$\Rightarrow dy=\frac{du}{{\sec }^{2}y}$
$\int \frac{{\sec }^{2}x}{u}\cdot \frac{du}{{\sec }^{2}x}=-\int \frac{{\sec }^{2}y}{v}\cdot \frac{dv}{{\sec }^{2}y}$
$\Rightarrow \int \frac{du}{u}=-\int \frac{dv}{v}$
$\Rightarrow \log |x|=-\log |v|+\log |c|$
$\Rightarrow \log |\tan x|=-\log |\tan y|+\log |c|$
$\Rightarrow \log |\tan x\cdot \tan y|=\log |c|$
$(\because \log m+\log n=\log mn)$
$\Rightarrow (\because \log m=\log n\Rightarrow m=n)$
which is the required general solution.