Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos 2x = 2{\cos ^2}x - 1 = 1 - 2{\sin ^2}x\\
\dfrac{{1 + \cos x}}{{1 - \cos x}}.ta{n^2}\dfrac{x}{2}\\
= \dfrac{{1 + \left( {2{{\cos }^2}\dfrac{x}{2} - 1} \right)}}{{1 - \left( {1 - 2{{\sin }^2}\dfrac{x}{2}} \right)}}.{\tan ^2}\dfrac{x}{2}\\
= \dfrac{{2{{\cos }^2}\dfrac{x}{2}}}{{2{{\sin }^2}\dfrac{x}{2}}}.\dfrac{{{{\sin }^2}\dfrac{x}{2}}}{{{{\cos }^2}\dfrac{x}{2}}}\\
= 1
\end{array}\)
