Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\cos 2x = {\cos ^2}x - {\sin ^2}x\\
\sin 2x = 2\sin x.\cos x\\
\Rightarrow \cot x - \tan x - 2\tan 2x\\
= \dfrac{{\cos x}}{{\sin x}} - \dfrac{{\sin x}}{{\cos x}} - \dfrac{{2\sin 2x}}{{\cos 2x}}\\
= \dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{\sin x.\cos x}} - \dfrac{{2\sin 2x}}{{\cos 2x}}\\
= \dfrac{{\cos 2x}}{{\dfrac{1}{2}\sin 2x}} - \dfrac{{2\sin 2x}}{{\cos 2x}}\\
= \dfrac{{2\cos 2x}}{{\sin 2x}} - \dfrac{{2\sin 2x}}{{\cos 2x}}\\
= 2.\dfrac{{{{\cos }^2}2x - {{\sin }^2}2x}}{{\sin 2x.\cos 2x}}\\
= 2.\dfrac{{\cos 4x}}{{\dfrac{1}{2}\sin 4x}}\\
= \dfrac{{4\cos 4x}}{{\sin 4x}}\\
= 4\cot 4x\\
c,\\
\dfrac{1}{{{{\cos }^6}x}} - {\tan ^6}x\\
= \dfrac{1}{{{{\cos }^6}x}} - \dfrac{{{{\sin }^6}x}}{{{{\cos }^6}x}}\\
= \dfrac{{1 - {{\sin }^6}x}}{{{{\cos }^6}x}}\\
= \dfrac{{\left( {1 - {{\sin }^2}x} \right)\left( {1 + {{\sin }^2}x + {{\sin }^4}x} \right)}}{{{{\cos }^6}x}}\\
= \dfrac{{{{\cos }^2}x.\left( {1 + {{\sin }^2}x + {{\sin }^4}x} \right)}}{{{{\cos }^6}x}}\\
= \dfrac{{1 + {{\sin }^2}x + {{\sin }^4}x}}{{{{\cos }^4}x}}\\
= \dfrac{{1 + {{\sin }^2}x + {{\left( {1 - {{\cos }^2}x} \right)}^2}}}{{{{\cos }^4}x}}\\
= \dfrac{{1 + {{\sin }^2}x + 1 - 2{{\cos }^2}x + {{\cos }^4}x}}{{{{\cos }^4}x}}\\
= \dfrac{{2\left( {1 - {{\cos }^2}x} \right) + {{\sin }^2}x + {{\cos }^4}x}}{{{{\cos }^4}x}}\\
= \dfrac{{3{{\sin }^2}x + {{\cos }^4}x}}{{{{\cos }^4}x}}\\
= \dfrac{{3.\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}}{{{{\cos }^2}x}} + 1\\
= \dfrac{{3{{\tan }^2}x}}{{{{\cos }^2}x}} + 1
\end{array}\)
