Question

2sinA+3cosA/5sinA-cosA
если ctgA= -2

Answer 1

$\frac{2Sin\alpha+3Cos\alpha}{5Sin\alpha -Cos\alpha}=\frac{\frac{2Sin\alpha}{Sin\alpha}+3\frac{Cos\alpha}{Sin\alpha}}{5\frac{Sin\alpha}{Sin\alpha}-\frac{Cos\alpha}{Sin\alpha}}=\frac{2+3Ctg\alpha}{5-Ctg\alpha}=\frac{2+3*(-2)}{5-(-2)}=\frac{2-6}{5+2}=-\frac{4}{7}\\\\Otvet:\boxed{-\frac{4}{7}}$