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

.: We have, $\cfrac{{dy}}{{dx}} = \cfrac{{1 - \cos x}}{{1 + \cos x}} \Rightarrow {\rm{ }}\cfrac{{dy}}{{dx}} = \cfrac{{2{{\sin }^2}\left( {\cfrac{x}{2}} \right)}}{{2{{\cos }^2}\left( {\cfrac{x}{2}} \right)}}$

$\Rightarrow \cfrac{{dy\prime }}{{dx}} = {\tan ^2}\left( {\cfrac{x}{2}} \right) \Rightarrow dy = {\tan ^2}\left( {\cfrac{x}{2}} \right)dx$
…(1)
Integrating (1) both sides,

we get
$\int d y = \int {{{\tan }^2}} \cfrac{x}{2}dx$

$\Rightarrow y = \int {\left( {{{\sec }^2}\cfrac{x}{2} - 1} \right)} dx \Rightarrow y = 2\tan \cfrac{x}{2} - x + C$

which is required solution.