Giải thích các bước giải:
ĐKXĐ:
Ta có:
\(\begin{array}{l}
\frac{{\sin x}}{{\sin x + \cos x}} - \frac{{\cos x}}{{\cos x - \sin x}} = \frac{{1 + {{\cot }^2}x}}{{1 - {{\cot }^2}x}}\\
\Leftrightarrow \frac{{\sin x\left( {\cos x - \sin x} \right) - \cos x\left( {\sin x + \cos x} \right)}}{{\left( {\sin x + \cos x} \right)\left( {\cos x - \sin x} \right)}} = \frac{{1 + \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}}}}{{1 - \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}}}}\\
\Leftrightarrow \frac{{\sin x.\cos x - {{\sin }^2}x - \cos x.\sin x - {{\cos }^2}x}}{{{{\cos }^2}x - {{\sin }^2}x}} = \frac{{{{\sin }^2}x + {{\cos }^2}}}{{{{\sin }^2}x}}:\frac{{{{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}}}\\
\Leftrightarrow \frac{{ - {{\sin }^2}x - {{\cos }^2}x}}{{{{\cos }^2}x - {{\sin }^2}x}} = \frac{1}{{{{\sin }^2}x - {{\cos }^2}x}}\\
\Leftrightarrow \frac{{ - 1}}{{{{\cos }^2}x - {{\sin }^2}x}} = \frac{1}{{{{\sin }^2}x - {{\cos }^2}x}}
\end{array}\)
Pt trên luôn đúng với mọi x