Докажите тождество пожалуйста

Решите задачу:

$\frac{cos^22a-4cos^2a+3}{cos^22a+4cos^2a-1} =tg^4a\\\\ \frac{cos^22a-4cos^2a+3sin^2a+3cos^2a}{cos^22a+4cos^2a-sin^2a-cos^2a} = \frac{sin^4a}{cos^4a} \\\\ \frac{cos^22a-cos^2a+3sin^2a}{cos^22a+3cos^2a-sin^2a} = \frac{sin^2a*sin^2a}{cos^2a*cos^2a} \\\\ \frac{ \frac{1+cos4a-1-cos2a+3-3cosa}{2} }{ \frac{1+cos4a+3+3cos2a-1+cos2a}{2} } = \frac{ \frac{(1-cos2a)^2}{4} }{ \frac{(1+cos2a)^2}{4} } \\\\ \frac{1+cos4a-1-cos2a+3-cos2a}{1+cos4a+3+3cos2a-1+cos2a} = \frac{1-cos2a+cos^22a}{1+2cos2a+cos^22a}$ $\frac{cos4a-2cos2a+3}{cos4a+4cos2a+3} = \frac{1-co2a+ \frac{1+cos4a}{2} }{1+2cos2a+ \frac{1+cos4a}{2}} \\\\ \frac{cos4a-2cos2a+3}{cos4a+4cos2a+3} = \frac{2-2cos2a+1+cos4a}{2+4cos2a+1+cos4a} \\\\ \frac{cos4a-2cos2a+3}{cos4a+4cos2a+3}=\frac{cos4a-2cos2a+3}{cos4a+4cos2a+3}\\\\1=1$