Question

$\frac{dy}{dx}=\frac{1-cosx}{1+cosx}$

Answer 1

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

We know that $\cos x=2{\cos }^2\frac{x}{2}-1=1-{\sin }^2\frac{x}{2}$

$\Rightarrow \frac{dy}{dx}=\frac{1-(1-{\sin }^2\frac{x}{2})}{1+(2{\cos }^2\frac{x}{2}-1)}$

$=\frac{1-1+{\sin }^2\frac{x}{2}}{1+2{\cos }^2\frac{x}{2}-1}$

$=\frac{{\sin }^2\frac{x}{2}}{2{\cos }^2\frac{x}{2}}$

$={\tan }^2\frac{x}{2}$

$={\sec }^2\frac{x}{2}-1$

$\Rightarrow \frac{dy}{dx}={\sec }^2\frac{x}{2}-1$

$\Rightarrow dy=({\sec }^2\frac{x}{2}-1)dx$

$\Rightarrow \int dy=\int ({\sec }^2\frac{x}{2}-1)dx$

$\Rightarrow \int dy=\int {\sec }^2\frac{x}{2}dx-\int dx$

$\Rightarrow y=\frac{\tan \frac{x}{2}}{\frac{1}{2}}-x+c$ where $c$ is the constant of integration

$\therefore y=2\tan \frac{x}{2}-x+c$where $c$ is the constant of integration