Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
0 < \alpha  < \dfrac{\pi }{2} \Rightarrow \left\{ \begin{array}{l}
\sin \alpha  > 0\\
cos\alpha  > 0
\end{array} \right.\\
{\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\\
\cos \alpha  > 0 \Rightarrow \cos \alpha  = \sqrt {1 - {{\sin }^2}\alpha }  = \dfrac{4}{5}\\
\tan \alpha  = \dfrac{{\sin \alpha }}{{\cos \alpha }} = \dfrac{3}{4};\,\,\,\,\,\cot \alpha  = \dfrac{{\cos \alpha }}{{\sin \alpha }} = \dfrac{4}{3}\\
\sin 2\alpha  = 2\sin \alpha .\cos \alpha  = \dfrac{{24}}{{25}}\\
\cos 2\alpha  = 2{\cos ^2}\alpha  - 1 = \dfrac{7}{{25}}\\
2.\\
A = \dfrac{{\sin 2x - \sin x}}{{1 - \cos x + \cos 2x}} = \dfrac{{2\sin x.\cos x - \sin x}}{{1 - \cos x + \left( {2{{\cos }^2}x - 1} \right)}}\\
 = \dfrac{{\sin x.\left( {2\cos x - 1} \right)}}{{\cos x\left( {2\cos x - 1} \right)}} = \dfrac{{\sin x}}{{\cos x}} = \tan x\\
B = \sin \left( {\dfrac{\pi }{4} + x} \right) - \cos \left( {\dfrac{\pi }{4} - x} \right)\\
 = \sin \dfrac{\pi }{4}.\cos x + \cos \dfrac{\pi }{4}.\sin x - \left( {\cos \dfrac{\pi }{4}.\cos x + \sin \dfrac{\pi }{4}.\sin x} \right)\\
 = \dfrac{{\sqrt 2 }}{2}\cos x + \dfrac{{\sqrt 2 }}{2}\sin x - \dfrac{{\sqrt 2 }}{2}\cos x - \dfrac{{\sqrt 2 }}{2}\sin x\\
 = 0\\
3,\\
\tan x + \cot x + \tan 3x + \cot 3x\\
 = \dfrac{{\sin x}}{{\cos x}} + \dfrac{{\cos x}}{{\sin x}} + \dfrac{{\sin 3x}}{{\cos 3x}} + \dfrac{{\cos 3x}}{{\sin 3x}}\\
 = \dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{\sin x.\cos x}} + \dfrac{{{{\sin }^2}3x + {{\cos }^2}3x}}{{\sin 3x.\cos 3x}}\\
 = \dfrac{1}{{\dfrac{1}{2}\sin 2x}} + \dfrac{1}{{\dfrac{1}{2}\sin 6x}}\\
 = \dfrac{2}{{\sin 2x}} + \dfrac{2}{{\sin 6x}}\\
 = \dfrac{{2.\left( {\sin 6x + \sin 2x} \right)}}{{\sin 2x.\sin 6x}}\\
 = \dfrac{{2.2.\sin \dfrac{{6x + 2x}}{2}.\cos \dfrac{{6x - 2x}}{2}}}{{\sin 2x.sin6x}}\\
 = \dfrac{{4\sin 4x.\cos 2x}}{{\sin 2x.\sin 6x}}\\
 = \dfrac{{8.\sin 2x.\cos 2x.\cos 2x}}{{\sin 2x.sin6x}}\\
 = \dfrac{{8{{\cos }^2}2x}}{{\sin 6x}}
\end{array}\)