если tg=-4

Решите задачу:

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