Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
sin\left( {a + b} \right) = \sin a.\cos b + \cos a.\sin b\\
A = \sin \left( {x + y} \right).\cos \left( {x - y} \right) + \sin \left( {x - y} \right).\cos \left( {x + y} \right)\\
 = \sin \left[ {\left( {x + y} \right) + \left( {x - y} \right)} \right] = \sin 2x\\
B = \sin 6^\circ .cos12^\circ .\cos 24^\circ .\cos 48^\circ \\
 = \frac{1}{{2\cos 6^\circ }}.\left( {2\sin 6^\circ .\cos 6^\circ } \right).cos12^\circ .\cos 24^\circ .\cos 48^\circ \\
 = \frac{1}{{2\cos 6^\circ }}.\sin 12^\circ .\cos 12^\circ .\cos 24^\circ .\cos 48^\circ \\
 = \frac{1}{{2\cos 6^\circ }}.\frac{1}{2}\sin 24^\circ .\cos 24^\circ .\cos 48^\circ \\
 = \frac{1}{{2\cos 6^\circ }}.\frac{1}{2}.\frac{1}{2}\sin 48^\circ .\cos 48^\circ \\
 = \frac{1}{{2\cos 6^\circ }}.\frac{1}{2}.\frac{1}{2}.\frac{1}{2}\sin 96^\circ \\
 = \frac{1}{{16}}.\frac{{\sin 96^\circ }}{{\cos 6^\circ }} = \frac{1}{{16}}.\frac{{\cos \left( {90^\circ  - 96^\circ } \right)}}{{\cos 6^\circ }} = \frac{1}{{16}}.\frac{{\cos \left( { - 6^\circ } \right)}}{{\cos 6^\circ }} = \frac{1}{{16}}\\
2,\\
\tan \alpha  - \cot \alpha  = 2\\
 \Leftrightarrow {\left( {\tan \alpha  - \cot \alpha } \right)^2} = 4\\
 \Leftrightarrow {\tan ^2}\alpha  - 2\tan \alpha .\cot \alpha  + {\cot ^2}\alpha  = 4\\
 \Leftrightarrow {\tan ^2}\alpha  + {\cot ^2}\alpha  - 2 = 4\\
 \Leftrightarrow \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} + \frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = 6\\
 \Leftrightarrow \frac{{{{\sin }^4}\alpha  + {{\cos }^4}\alpha }}{{{{\sin }^2}\alpha .{{\cos }^2}\alpha }} = 6\\
 \Leftrightarrow \frac{{{{\left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right)}^2} - 2{{\sin }^2}\alpha .{{\cos }^2}\alpha }}{{{{\sin }^2}a.{{\cos }^2}\alpha }} = 6\\
 \Leftrightarrow \frac{1}{{{{\sin }^2}a.{{\cos }^2}\alpha }} - 2 = 6\\
 \Leftrightarrow {\sin ^2}a.{\cos ^2}\alpha  = \frac{1}{8}\\
A = \frac{1}{{{{\sin }^2}\alpha }} + \frac{1}{{{{\cos }^2}\alpha }} = \frac{{{{\sin }^2}\alpha  + {{\cos }^2}\alpha }}{{{{\sin }^2}\alpha .co{s^2}\alpha }} = \frac{1}{{{{\sin }^2}\alpha .{{\cos }^2}\alpha }} = 8
\end{array}\)