Giải thích các bước giải:
\(\begin{array}{l}
E = \frac{1}{{{{\cos }^6}x}} - {\tan ^6}x - \frac{{3{{\tan }^2}x}}{{{{\cos }^2}x}}\\
= \frac{1}{{{{\cos }^6}x}} - \frac{{{{\sin }^6}x}}{{{{\cos }^6}x}} - \frac{{3.\frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}}{{{{\cos }^2}x}}\\
= \frac{{1 - {{\sin }^6}x}}{{{{\cos }^6}x}} - 3.\frac{{{{\sin }^2}x}}{{{{\cos }^4}x}}\\
= \frac{{\left( {1 - {{\sin }^2}x} \right)\left( {1 + {{\sin }^2}x + {{\sin }^4}x} \right)}}{{{{\cos }^6}x}} - \frac{{3{{\sin }^2}x}}{{{{\cos }^4}x}}\\
= \frac{{{{\cos }^2}x.\left( {1 + {{\sin }^2}x + {{\sin }^4}x} \right)}}{{{{\cos }^6}x}} - \frac{{3{{\sin }^2}x}}{{{{\cos }^4}x}}\\
= \frac{{1 + {{\sin }^2}x + {{\sin }^4}x - 3{{\sin }^2}x}}{{{{\cos }^4}x}}\\
= \frac{{{{\left( {1 - {{\sin }^2}x} \right)}^2}}}{{{{\cos }^4}x}}\\
= \frac{{{{\left( {{{\cos }^2}x} \right)}^2}}}{{{{\cos }^4}x}}\\
= 1\\
F = \frac{{\left( {{{\sin }^2}x + {{\tan }^2}x + 1} \right)\left( {{{\cos }^2}x - {{\cot }^2}x + 1} \right)}}{{\left( {{{\cos }^2}x + {{\cot }^2}x + 1} \right)\left( {{{\sin }^2}x - {{\tan }^2}x + 1} \right)}}\\
= \frac{{\left( {{{\sin }^2}x + \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + 1} \right)\left( {{{\cos }^2}x - \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}} + 1} \right)}}{{\left( {{{\cos }^2}x + \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}} + 1} \right)\left( {{{\sin }^2}x - \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + 1} \right)}}\\
= \frac{{\left( {{{\sin }^2}x{{\cos }^2}x + {{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\cos }^2}x{{\sin }^2}x - {{\cos }^2}x + {{\sin }^2}x} \right)}}{{\left( {{{\cos }^2}x{{\sin }^2}x + {{\cos }^2}x + {{\sin }^2}x} \right)\left( {{{\sin }^2}x{{\cos }^2}x - {{\sin }^2}x + {{\cos }^2}x} \right)}}\\
= \frac{{{{\sin }^2}x{{\cos }^2}x\left( {{{\cos }^2}x{{\sin }^2}x - {{\cos }^2}x + {{\sin }^2}x} \right)}}{{{{\cos }^2}x{{\sin }^2}x.\left( {{{\sin }^2}x{{\cos }^2}x - {{\sin }^2}x + {{\cos }^2}x} \right)}}\\
= \frac{{{{\cos }^2}x{{\sin }^2}x - {{\cos }^2}x + {{\sin }^2}x}}{{{{\sin }^2}x{{\cos }^2}x - {{\sin }^2}x + {{\cos }^2}x}}\\
= \frac{{{{\cos }^2}x\left( {1 - {{\cos }^2}x} \right) - {{\cos }^2}x + 1 - {{\cos }^2}x}}{{{{\cos }^2}x\left( {1 - {{\cos }^2}x} \right) - \left( {1 - {{\cos }^2}x} \right) + {{\cos }^2}x}}\\
= \frac{{ - {{\cos }^4}x - {{\cos }^2}x + 1}}{{ - {{\cos }^4}x + 3{{\cos }^2}x - 1}}
\end{array}\)
