Q. The general solution of the differential equation $\frac{dy}{dx} + \frac{1 + \cos{2y}}{1 - \cos{2x}} = 0$ is given by
A) $\tan y + \cot x = c$
B) $\tan y - \cot x = c$
C) $\tan x - \cot y = c$
D) $\tan x + \cot y = c$
Explanation
(b)
We have,
$\frac{dy}{dx} + \frac{1 + \cos{2y}}{1 - \cos{2x}} = 0$
Given, $\frac{dy}{dx} = - \frac{1 + \cos{2y}}{1 - \cos{2x}} = - \frac{2\cos^{2}y}{2\sin^{2}x}$
$\Rightarrow$ $\int_{}^{}{\sec^{2}y}\ dy = - \int_{}^{}{cosec^{2}x}\ dx$
$\Rightarrow \tan{y = \cot{x + c.}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.