Toán Sinx/(sinx+cosx) – cosx/(cosx-sinx)=1+cotx^2/1-cotx^2 21/07/2021 By Liliana Sinx/(sinx+cosx) – cosx/(cosx-sinx)=1+cotx^2/1-cotx^2
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 Trả lời