Question

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

Answer 1

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