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