Question

\[\frac{dy}{dx} = \frac{1 - \cos 2y}{1 + \cos 2y}\]

Answer 1

We have,
\[\frac{dy}{dx} = \frac{1 - \cos 2y}{1 + \cos 2y}\]
\[ \Rightarrow \frac{dx}{dy} = \frac{1 + \cos 2y}{1 - \cos 2y}\]
\[ \Rightarrow dx = \frac{1 + \cos 2y}{1 - \cos 2y}dy\]
\[ \Rightarrow dx = \frac{2 \cos^2 y}{2 \sin^2 y}dy\]
\[ \Rightarrow dx = \cot^2 y\ dy\]
Integrating both sides, we get
\[ \Rightarrow \int dx = \int \cot^2 y\ dy\]
\[ \Rightarrow x = \int\left( {cosec}^2 y - 1 \right) dy\]
\[ \Rightarrow x = \int {cosec}^2 y dy - \int dy\]
\[ \Rightarrow x = - \cot y - y + C\]
\[ \Rightarrow x + \cot y + y = C\]
\[\text{Hence, }x + \cot y + y =\text{C is the required solution.}\]