Question

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

Answer 1

We have, 
\[\frac{dy}{dx} = \frac{1 - \cos x}{1 + \cos x}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{2 \sin^2 \frac{x}{2}}{2 \cos^2 \frac{x}{2}}\]
\[ \Rightarrow \frac{dy}{dx} = \tan^2 \frac{x}{2}\]
\[ \Rightarrow dy = \left( \tan^2 \frac{x}{2} \right)dx\]
Integrating both sides, we get
\[\int dy = \int\left( \tan^2 \frac{x}{2} \right)dx\]
\[ \Rightarrow \int dy = \int\left( \sec^2 \frac{x}{2} - 1 \right)dx\]
\[ \Rightarrow y = 2 \tan \frac{x}{2} - x + C\]
\[\text{ So, } y = 2 \tan \frac{x}{2} - x + \text{ C is defined for all }x \in R . \]
\[\text{ Hence, }y = 2 \tan \frac{x}{2} - x +\text{ C, where }x \in R, \text{ is the solution to the given differential equation } .\]