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