Giải thích các bước giải:

\(\begin{array}{l}
1,\\
\frac{{\sin \left( {\alpha  + \beta } \right) + \sin \left( {\alpha  - \beta } \right)}}{{\cos \left( {\alpha  + \beta } \right) - \cos \left( {\alpha  - \beta } \right)}} = \frac{{2.\sin \frac{{\alpha  + \beta  + \alpha  - \beta }}{2}.\cos \frac{{\alpha  + \beta  - \alpha  + \beta }}{2}}}{{ - 2.\sin \frac{{\alpha  + \beta  + \alpha  - \beta }}{2}.\sin \frac{{\alpha  + \beta  - \alpha  + \beta }}{2}}} = \frac{{2\sin \alpha .\cos \beta }}{{ - 2\sin \alpha .sin\beta }} = \frac{{ - \cos \beta }}{{\sin \beta }} =  - \cot \beta \\
2,\\
\frac{{\cos \left( {45^\circ  - a} \right) - \cos \left( {45^\circ  + a} \right)}}{{\sin \left( {45^\circ  - a} \right) - \sin \left( {45^\circ  - a} \right)}} = \frac{{ - 2.sin\frac{{45^\circ  - a + 45^\circ  + a}}{2}.\sin \frac{{45^\circ  - a - 45^\circ  - a}}{2}}}{{2\cos \frac{{45^\circ  + a + 45^\circ  - a}}{2}.sin\frac{{45^\circ  + a - 45^\circ  + a}}{2}}} = \frac{{ - 2\sin 45^\circ .\sin \left( { - a} \right)}}{{2.\cos 45^\circ .\sin a}} = \frac{{2\sin 45^\circ .\sin a}}{{2\cos 45^\circ .\sin a}} = 1\\
3,\\
\sin \left( {2x - \pi } \right).\cos \left( {x - 3\pi } \right) + \sin \left( {2x - \frac{{9\pi }}{2}} \right).\cos \left( {x + \frac{\pi }{2}} \right)\\
 = \sin \left( {\pi  - \left( {2x - \pi } \right)} \right).\left( { - \cos \left( {\pi  - \left( {x - 3\pi } \right)} \right)} \right) + \sin \left( {2x - \frac{\pi }{2} - 4\pi } \right).\cos \left( {x + \frac{\pi }{2}} \right)\\
 =  - \sin \left( {2\pi  - 2x} \right).\cos \left( {4\pi  - x} \right) + \sin \left( {2x - \frac{\pi }{2}} \right).sin\left( {\frac{\pi }{2} - \left( {x + \frac{\pi }{2}} \right)} \right)\\
 =  - \sin \left( { - 2x} \right).\cos \left( { - x} \right) + \cos \left( {\frac{\pi }{2} - \left( {2x - \frac{\pi }{2}} \right)} \right).\sin \left( { - x} \right)\\
 = \sin 2x.\cos x + \cos \left( {\pi  - 2x} \right).\left( { - \sin x} \right)\\
 = \sin 2x.\cos x + \left( { - \cos 2x} \right).\left( { - \sin x} \right)\\
 = \sin 2x.\cos x + \cos 2x.\sin x\\
 = \sin \left( {2x + x} \right) = \sin 3x
\end{array}\)