\(\left(cosa-sina\right)^2=\frac{1}{25}\Leftrightarrow sin^2a+cos^2a-2sina.cosa=\frac{1}{25}\)
\(\Leftrightarrow\frac{sin^2a+cos^2a-2sina.cosa}{sin^2a}=\frac{1}{5sin^2a}=\frac{sin^2a+cos^2a}{5sin^2a}\)
\(\Leftrightarrow1+cot^2a-2cota=\frac{1}{5}+\frac{1}{5}cot^2a\)
\(\Leftrightarrow4cot^2a-10cota+4=0\Rightarrow\left[{}\begin{matrix}cota=2\\cota=\frac{1}{2}\end{matrix}\right.\)
Thui vậy! OK anh e sẽ giúp! Mà hok trc lp 9 hay sao mà chăm dữ?!
Có \(\cos\alpha-\sin\alpha=\frac{1}{5}\Rightarrow\left(\cos\alpha-\sin\alpha\right)^2=\frac{1}{25}\)
\(\Leftrightarrow\cos^2\alpha-2\sin\alpha.\cos\alpha+\sin^2\alpha=\frac{1}{25}\)
\(\Leftrightarrow1-2\sin\alpha.\cos\alpha=\frac{1}{25}\)
\(\Leftrightarrow\sin\alpha.\cos\alpha=\frac{12}{25}\Leftrightarrow\sin\alpha=\frac{12}{25\cos\alpha}\)
Thay vào biểu thức ban đầu rùi giải pt b2 là OK
