По формуле двойного аргумента....

Решите задачу:

$\dfrac{sin^22a-4sin^2a}{sin^22a+4sin^2a-4}= \\ \\ = \dfrac{sin^22a-4sin^2a}{sin^22a+4(-(1-sin^2a))} = \\ \\ = \dfrac{sin^22a-4sin^2a}{sin^22a-4cos^2a}= \\ \\ = \dfrac{4sin^2acos^2a-4sin^2a}{4sin^2acos^2a-4cos^2a}= \\ \\ = \dfrac{4sin^2a(-(1-cos^2a))}{4cos^2a(-(1-sin^2a))}= \\ \\ = \dfrac{sin^4a}{cos^4a}= \\ \\ =tg^4a$