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