Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
\dfrac{{2\cos x}}{{\cot \dfrac{x}{2} - \tan \dfrac{x}{2}}} = \dfrac{{2\cos x}}{{\dfrac{{\cos \dfrac{x}{2}}}{{\sin \dfrac{x}{2}}} - \dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}}} = \dfrac{{2\cos x}}{{\dfrac{{{{\cos }^2}\dfrac{x}{2} - {{\sin }^2}\dfrac{x}{2}}}{{\sin \dfrac{x}{2}.\cos \dfrac{x}{2}}}}} = \dfrac{{2\cos x}}{{\dfrac{{\cos x}}{{\dfrac{1}{2}\sin x}}}} = \dfrac{{2\cos x.\dfrac{1}{2}\sin x}}{{\cos x}} = \sin x\\
b,\\
\sin \dfrac{A}{2}.{\cos ^3}\dfrac{B}{2} = \sin \dfrac{B}{2}.{\cos ^3}\dfrac{A}{2}\\
 \Leftrightarrow \sin \dfrac{A}{2}.\cos \dfrac{B}{2}.\left( {1 - {{\sin }^2}\dfrac{B}{2}} \right) = \sin \dfrac{B}{2}.\cos \dfrac{A}{2}.\left( {1 - {{\sin }^2}\dfrac{A}{2}} \right)\\
 \Leftrightarrow \sin \dfrac{A}{2}.\cos \dfrac{B}{2} - \sin \dfrac{A}{2}.\cos \dfrac{B}{2}.{\sin ^2}\dfrac{B}{2} = \sin \dfrac{B}{2}.\cos \dfrac{A}{2} - \sin \dfrac{B}{2}.\cos \dfrac{A}{2}.{\sin ^2}\dfrac{A}{2}\\
 \Leftrightarrow \left( {\sin \dfrac{A}{2}.\cos \dfrac{B}{2} - \sin \dfrac{B}{2}.\cos \dfrac{A}{2}} \right) - \sin \dfrac{A}{2}.sin\dfrac{B}{2}\left( {\sin \dfrac{B}{2}.\cos \dfrac{B}{2} - \sin \dfrac{A}{2}.\cos \dfrac{A}{2}} \right) = 0\\
 \Leftrightarrow \sin \left( {\dfrac{A}{2} - \dfrac{B}{2}} \right) - \sin \dfrac{A}{2}.\sin \dfrac{B}{2}.\left( {\dfrac{1}{2}\sin B - \dfrac{1}{2}\sin A} \right) = 0\\
 \Leftrightarrow \sin \dfrac{{A - B}}{2} - \sin \dfrac{A}{2}.\sin \dfrac{B}{2}.\dfrac{1}{2}.2.\cos \dfrac{{B + A}}{2}.\sin \dfrac{{B - A}}{2} = 0\\
 \Leftrightarrow \sin \dfrac{{A - B}}{2}\left( {1 + \sin \dfrac{A}{2}.\sin \dfrac{B}{2}.\cos \dfrac{{B + A}}{2}} \right) = 0\\
 \Rightarrow \sin \dfrac{{A - B}}{2} = 0\\
 \Leftrightarrow \dfrac{{A - B}}{2} = 0 \Leftrightarrow \widehat A = \widehat B\\
c,\\
A = 2\sin x.\cos 3x + 2{\sin ^2}\left( {x + \dfrac{\pi }{4}} \right) - \sin 4x\\
 = \left[ {\sin \left( {x + 3x} \right) + \sin \left( {x - 3x} \right)} \right] + 2.\left( {\sin x.\cos \dfrac{\pi }{4} + \cos x.sin\dfrac{\pi }{4}} \right) - \sin 4x\\
 = \sin 4x + \sin \left( { - 2x} \right) + 2.{\left( {\dfrac{{\sqrt 2 }}{2}.\sin x + \dfrac{{\sqrt 2 }}{2}\cos x} \right)^2} - \sin 4x\\
 =  - \sin 2x + 2.{\left( {\dfrac{{\sqrt 2 }}{2}} \right)^2}.{\left( {\sin x + \cos x} \right)^2}\\
 =  - \sin 2x + \left( {{{\sin }^2}x + 2\sin x.\cos x + {{\cos }^2}x} \right)\\
 =  - 2\sin x.\cos x + \left( {1 + 2\sin x.\cos x} \right)\\
 = 1
\end{array}\)