Question

$\frac{3cos \alpha + sin \alpha}{2cos - sin \alpha }$

ctgA=2
с решением ​

Answer 1

$\displaystyle\bf\\Ctg\alpha =2\\\\Ctg\alpha =\frac{Cos\alpha }{Sin\alpha } \\\\\frac{Cos\alpha }{Sin\alpha } =2 \ \ \ \Rightarrow \ \ \ Cos\alpha =2Sin\alpha \\\\\\\frac{3Cos\alpha +Sin\alpha }{2Cos\alpha -Sin\alpha } =\frac{3\cdot 2Sin\alpha +Sin\alpha }{2\cdot 2Sin\alpha -Sin\alpha } =\frac{6Sin\alpha +Sin\alpha }{4Sin\alpha -Sin\alpha } =\\\\\\=\frac{7Sin\alpha }{3Sin\alpha } =\frac{7}{3} =2\frac{1}{3}$