Question

$frac{3cos2alpha} {2-9cos^2alpha }$

если tg=-4

Answer 1

$tga=-4\\frac{3, cos2a}{2-9, cos^2a}=frac{3cdot (2cos^2a-1)}{2-9, cos^2a}=frac{3, cdot cos^2a(2-frac{1}{cos^2a})}{cos^2acdot (frac{2}{cos^2a}-9)}=frac{3cdot (2-(1+tg^2a))}{2cdot (1+tg^2a)-9}=\\=frac{3cdot (2-(1-4))}{2cdot (1-4)-9}=frac{3cdot 5}{-2cdot 3-9}=frac{15}{-15}=-1\\\star ; ; 1+tg^2a=frac{1}{cos^2a}; ; ,; ; cos2a=cos^2a-sin^2a=2cos^2a-1; ; star$