Question

(sin^2a- cos^2a + cos^4a)/(cos^2a - sin^2a + sin^4a)=tg^4a

Answer 1

$\frac{1-cos^2a-cos^2a+cos^4a}{1-sin^2a-sin^2a+sin^4a}=tg^4a\\ \frac{(1-cos^2a)-cos^2a(1-cos^2a)}{(1-sin^2a)-sin^2a(1-sin^2a)}=tg^4a\\ \frac{(1-cos^2a)(1-cos^2a)}{(1-sin^2a)(1-sin^2a)}=tg^4a\\ \frac{(1-cos^2a)^2}{(1-sin^2a)^2}=tg^4a\\ \frac{sin^4a}{cos^4a}=tg^4a$

или по-другому:

$\frac{sin^2a-cos^2a+cos^4a}{cos^2a-sin^2a+sin^4a}=tg^4a\\ \frac{sin^2a-cos^2a(1-cos^a)}{cos^2a-sin^2a(1-sin^2a)}= \frac{sin^2a-cos^2asin^2a}{cos^2a-sin^2acos^2a}= \frac{sin^2a(1-cos^2a)}{cis^2a(1-sin^2a)}= \frac{sin^2a*sin^2a}{cos^2a*cos^2a}= \\ \frac{sin^4a}{cos^4a}=tg^4a$