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)

$\tan x.\tan y=k$

$\tan y.{\sec }^{2}xdx+\tan x.{\sec }^{2}ydy=0⟹\tan y{\sec }^{2}xdx=-\tan x{\sec }^{y}dy⟹\frac{{\sec }^{2}xdx}{\tan x}=-\frac{{\sec }^{2}ydy}{\tan y}$Integrating,$\int \frac{{\sec }^{2}xdx}{\tan x}=-\int \frac{{\sec }^{2}ydy}{\tan y}⟹\int \frac{d\left(\tan x\right)}{tanx}=-\int \frac{d\left(\tan y\right)}{\tan y}⟹\ln |\tan x|=-\ln |\tan y|+C⟹\ln |\tan x|+\ln |\tan y|=C⟹\ln |\tan x.\tan y|=C⟹\tan x.\tan y=e^C⟹\tan x.\tan y=k$