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

The correct option is D $\tan x.\tan y=k$

$\tan y{\sec }^{2}xdx+\tan x{\sec }^{2}ydy=0\Rightarrow \tan y{\sec }^{2}xdx=-\tan x{\sec }^{2}ydy\Rightarrow \frac{{\sec }^{2}x}{\tan x}dx=-\frac{{\sec }^{2}y}{\tan y}dy\Rightarrow \int \frac{{\sec }^{2}x}{\tan x}dx=\int -\frac{{\sec }^{2}y}{\tan y}dy$

let tanx be equal to b and tany equal to a

$\Rightarrow \int \frac{dx}{b}=-\int \frac{dy}{a}\Rightarrow \ln b=-\ln a+k\Rightarrow \ln ab=k\Rightarrow \ln \tan x\tan y=k\Rightarrow \tan x\tan y=k$