Câu a thử cho $x=60$ thấy hai vế không bằng nhau ($VT=-8, VP=8$)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
*)\\
1 - {\tan ^4}x = 1 - \dfrac{{{{\sin }^4}x}}{{{{\cos }^4}x}} = \dfrac{{{{\cos }^4}x - {{\sin }^4}x}}{{{{\cos }^4}x}}\\
 = \dfrac{{\left( {{{\cos }^2}x - {{\sin }^2}x} \right)\left( {{{\cos }^2}x + {{\sin }^2}x} \right)}}{{{{\cos }^4}x}}\\
 = \dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^4}x}}\\
 = \dfrac{{{{\cos }^2}x - \left( {1 - {{\cos }^2}x} \right)}}{{{{\cos }^4}x}}\\
 = \dfrac{{2{{\cos }^2}x - 1}}{{{{\cos }^4}x}}\\
 = \dfrac{2}{{{{\cos }^2}x}} - \dfrac{1}{{{{\cos }^4}x}}\\
*)\\
2{\cos ^2}x - \cos \left( {2x + \dfrac{\pi }{3}} \right) - 2\sin x.\cos \left( {x + \dfrac{\pi }{6}} \right)\\
 = 2{\cos ^2}x - \left( {\cos 2x.cos\dfrac{\pi }{3} - \sin 2x.\sin \dfrac{\pi }{3}} \right) - 2.\sin x.\left( {\cos x.\cos \dfrac{\pi }{6} - \sin x.\sin \dfrac{\pi }{6}} \right)\\
 = 2{\cos ^2}x - \left( {\dfrac{1}{2}\cos 2x - \dfrac{{\sqrt 3 }}{2}\sin 2x} \right) - 2\sin x\left( {\dfrac{{\sqrt 3 }}{2}\cos x - \dfrac{1}{2}\sin x} \right)\\
 = 2{\cos ^2}x - \dfrac{1}{2}\cos 2x + \dfrac{{\sqrt 3 }}{2}\sin 2x - \dfrac{{\sqrt 3 }}{2}.2\sin x.\cos x + {\sin ^2}x\\
 = 2{\cos ^2}x - \dfrac{1}{2}\left( {{{\cos }^2}x - {{\sin }^2}x} \right) + \dfrac{{\sqrt 3 }}{2}\sin 2x - \dfrac{{\sqrt 3 }}{2}\sin 2x + {\sin ^2}x\\
 = 2{\cos ^2}x - \dfrac{1}{2}{\cos ^2}x + \dfrac{1}{2}{\sin ^2}x + {\sin ^2}x\\
 = \dfrac{3}{2}\left( {{{\cos }^2}x + {{\sin }^2}x} \right)\\
 = \dfrac{3}{2}.1 = \dfrac{3}{2}
\end{array}\)