Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos x + \cos y = 2\cos \dfrac{{x + y}}{2}.\cos \dfrac{{x - y}}{2}\\
\cos x - \cos y = - 2\sin \dfrac{{x + y}}{2}.\sin \dfrac{{x - y}}{2}\\
\sin x + \sin y = 2\sin \dfrac{{x + y}}{2}.\cos \dfrac{{x - y}}{2}\\
\sin x - \sin y = 2.\cos \dfrac{{x + y}}{2}.\sin \dfrac{{x - y}}{2}\\
A = \dfrac{{\cos 7x - \cos 8x - \cos 9x + \cos 10x}}{{\sin 7x - \sin 8x - \sin 9x + \sin 10x}}\\
= \dfrac{{ - \left( {\cos 9x - \cos 7x} \right) + \left( {\cos 10x - \cos 8x} \right)}}{{ - \left( {\sin 9x - \sin 7x} \right) + \left( {\sin 10x - \sin 8x} \right)}}\\
= \dfrac{{2.sin8x.\sin x - 2\sin 9x.\sin x}}{{ - 2.\cos 8x.\sin x + 2.\cos 9x.\sin x}}\\
= \dfrac{{2\sin x.\left( {\sin 8x - \sin 9x} \right)}}{{2\sin x.\left( {\cos 9x - \cos 8x} \right)}}\\
= \dfrac{{\sin 8x - \sin 9x}}{{\cos 9x - \cos 8x}}\\
= \dfrac{{2.cos\dfrac{{17x}}{2}.\sin \dfrac{{ - x}}{2}}}{{ - 2\sin \dfrac{{17x}}{2}.\sin \dfrac{x}{2}}}\\
= \dfrac{{\cos \dfrac{{17x}}{2}}}{{\sin \dfrac{{17x}}{2}}}\\
= \cot \dfrac{{17x}}{2}\\
B = \dfrac{{\sin 2x + 2\sin 3x + \sin 4x}}{{\sin 3x + 2\sin 4x + \sin 5x}}\\
= \dfrac{{\left( {\sin 2x + \sin 4x} \right) + 2\sin 3x}}{{\left( {\sin 3x + \sin 5x} \right) + 2\sin 4x}}\\
= \dfrac{{2.\sin 3x.\cos x + 2\sin 3x}}{{2\sin 4x.\cos x + 2\sin 4x}}\\
= \dfrac{{2\sin 3x.\left( {\cos x + 1} \right)}}{{2\sin 4x.\left( {\cos x + 1} \right)}}\\
= \dfrac{{\sin 3x}}{{\sin 4x}}\\
C = \dfrac{{1 + \cos x + \cos 2x + \cos 3x}}{{\cos x + 2{{\cos }^2}x - 1}}\\
= \dfrac{{\left( {1 + \cos 2x} \right) + \left( {\cos x + \cos 3x} \right)}}{{\cos x + \left( {2{{\cos }^2}x - 1} \right)}}\\
= \dfrac{{\left( {1 + 2{{\cos }^2}x - 1} \right) + 2.\cos 2x.\cos x}}{{\cos x + \cos 2x}}\\
= \dfrac{{2\cos x\left( {\cos x + \cos 2x} \right)}}{{\cos x + \cos 2x}}\\
= 2\cos x\\
D = \dfrac{{\sin 4x + \sin 5x + \sin 6x}}{{\cos 4x + \cos 5x + \cos 6x}}\\
= \dfrac{{\left( {\sin 4x + \sin 6x} \right) + \sin 5x}}{{\left( {\cos 4x + \cos 6x} \right) + \cos 5x}}\\
= \dfrac{{2.sin5x.\cos x + \sin 5x}}{{2.\cos 5x.\cos x + \cos 5x}}\\
= \dfrac{{\sin 5x\left( {2\cos x + 1} \right)}}{{\cos 5x\left( {2\cos x + 1} \right)}}\\
= \dfrac{{\sin 5x}}{{\cos 5x}}\\
= \tan 5x
\end{array}\)
