Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
C = \dfrac{{1 - 2\cos 2x + \cos 4x}}{{1 + 2\cos 2x + \cos 4x}}\\
= \dfrac{{1 - 2\cos 2x + \left( {2{{\cos }^2}2x - 1} \right)}}{{1 + 2\cos 2x + \left( {2{{\cos }^2}2x - 1} \right)}}\\
= \dfrac{{2{{\cos }^2}2x - 2\cos 2x}}{{2{{\cos }^2}2x + 2\cos 2x}}\\
= \dfrac{{\cos 2x\left( {\cos 2x - 1} \right)}}{{\cos 2x\left( {\cos 2x + 1} \right)}}\\
= \dfrac{{\cos 2x - 1}}{{\cos 2x + 1}}\\
= \dfrac{{\left( {1 - 2{{\sin }^2}x} \right) - 1}}{{\left( {2{{\cos }^2}x - 1} \right) + 1}}\\
= \dfrac{{ - {{\sin }^2}x}}{{{{\cos }^2}x}} = - {\tan ^2}x\\
D = \left( {\dfrac{{1 + \cos \beta }}{{1 - \cos \beta }}} \right).{\tan ^2}\dfrac{\beta }{2} + {\sin ^2}\beta - 2\\
= \left( {\dfrac{{1 + \left( {2{{\cos }^2}\dfrac{\beta }{2} - 1} \right)}}{{1 - \left( {1 - 2{{\sin }^2}\dfrac{\beta }{2}} \right)}}} \right).\dfrac{{{{\sin }^2}\dfrac{\beta }{2}}}{{{{\cos }^2}\dfrac{\beta }{2}}} + {\sin ^2}\beta - 2\\
= \dfrac{{2{{\cos }^2}\dfrac{\beta }{2}}}{{2{{\sin }^2}\dfrac{\beta }{2}}}.\dfrac{{{{\sin }^2}\dfrac{\beta }{2}}}{{{{\cos }^2}\dfrac{\beta }{2}}} + {\sin ^2}\beta - 2\\
= 1 + {\sin ^2}\beta - 2\\
= {\sin ^2}\beta - 1 = - {\cos ^2}\beta
\end{array}\)
