ĐKXĐ:...
Ta có :[tex]tan^{2}x=tan^{2}\left | x \right |[/tex]
[tex]pt\Leftrightarrow tan^{2}\left | x \right |=\frac{1+cos\left | x \right |}{1+sin\left | x \right |}\Leftrightarrow sin^{2}\left | x \right |(1+sin\left | x \right |)=cos^{2}\left | x \right |(1+cos\left | x \right |)\Leftrightarrow sin^{2}\left | x \right |-cos^{2}\left | x \right |+sin^{3}\left | x \right |-cos^{3}\left | x \right |=0\Leftrightarrow (sin\left | x \right |-cos\left | x \right |)(sin\left | x \right |+cos\left | x \right |+sin^{2}\left | x \right |+sin\left | x \right |cos\left | x \right |+cos^{2}\left | x \right |)=0\Leftrightarrow (sin\left | x \right |-cos\left | x \right |)(1+sin\left | x \right |+cos\left | x \right |+sin\left | x \right |cos\left | x \right |)=0[/tex]
<=>...