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