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