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