Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
14,\\
a,\\
1 - {\cot ^4}x = 1 - \frac{{{{\cos }^4}x}}{{{{\sin }^4}x}} = \frac{{{{\sin }^4}x - {{\cos }^4}x}}{{{{\sin }^4}x}} = \frac{{\left( {{{\sin }^2}x - {{\cos }^2}x} \right)\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}}{{{{\sin }^4}x}} = \frac{{{{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^4}x}} = \frac{{{{\sin }^2}x - \left( {1 - {{\sin }^2}x} \right)}}{{{{\sin }^4}x}} = \frac{{2{{\sin }^2}x - 1}}{{{{\sin }^4}x}} = \frac{2}{{{{\sin }^2}x}} - \frac{1}{{{{\sin }^4}x}}\\
b,\\
\frac{{1 - 2\sin x.\cos x}}{{{{\cos }^2}x - {{\sin }^2}x}} = \frac{{{{\sin }^2}x + {{\cos }^2}x - 2\sin x.\cos x}}{{\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right)}} = \frac{{{{\left( {\sin x - \cos x} \right)}^2}}}{{\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right)}} = \frac{{\cos x - \sin x}}{{\cos x + \sin x}} = \frac{{1 - \frac{{\sin x}}{{\cos x}}}}{{1 + \frac{{\sin x}}{{\cos x}}}} = \frac{{1 - \tan x}}{{1 + \tan x}}\\
15,\\
a,\\
A = 3\left( {{{\sin }^4}x + {{\cos }^4}x} \right) - 2\left( {{{\sin }^6}x + {{\cos }^6}x} \right)\\
= 3\left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 2{{\sin }^2}x.{{\cos }^2}x} \right] - 2.\left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^3} - 3.{{\sin }^2}x.{{\cos }^2}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \right]\\
= 3\left[ {{1^2} - 2{{\sin }^2}x.{{\cos }^2}x} \right] - 2.\left[ {{1^3} - 3.{{\sin }^2}x.{{\cos }^2}x.1} \right]\\
= 3 - 6{\sin ^2}x.{\cos ^2}x - 2 + 6{\sin ^2}x.{\cos ^2}x\\
= 1\\
b,\\
B = \frac{{{{\left( {1 - {{\tan }^2}x} \right)}^2}}}{{4{{\tan }^2}x}} - \frac{1}{{4{{\sin }^2}x.{{\cos }^2}x}}\\
= \frac{{1 - 2{{\tan }^2}x + {{\tan }^4}x}}{{4{{\tan }^2}x}} - \frac{1}{{4{{\sin }^2}x.{{\cos }^2}x}}\\
= \frac{1}{{4{{\tan }^2}x}} - \frac{1}{2} + \frac{1}{4}{\tan ^2}x - \frac{1}{{4{{\sin }^2}x.{{\cos }^2}x}}\\
= \frac{{{{\cos }^2}x}}{{4{{\sin }^2}x}} - \frac{1}{2} + \frac{{{{\sin }^2}x}}{{4{{\cos }^2}x}} - \frac{1}{{4{{\sin }^2}x.{{\cos }^2}x}}\\
= \frac{{{{\cos }^4}x + {{\sin }^4}x - 1}}{{4{{\sin }^2}x.{{\cos }^2}x}} - \frac{1}{2}\\
= \frac{{{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 2{{\sin }^2}x.{{\cos }^2}x - 1}}{{4{{\sin }^2}x.{{\cos }^2}x}} - \frac{1}{2}\\
= \frac{{{1^2} - 2{{\sin }^2}x.{{\cos }^2}x - 1}}{{4{{\sin }^2}x.{{\cos }^2}x}} - \frac{1}{2}\\
= \frac{{ - 1}}{2} - \frac{1}{2} = - 1
\end{array}\)
