Question

Решите пожалуйста sin^4b-sin^6b/cos^4b-cos^6b

Answer 1

$dfrac{sin^4beta-sin^6beta}{cos^4beta-cos^6beta}=dfrac{sin^4beta(1-sin^2beta)}{cos^4beta(1-cos^2beta)}=dfrac{sin^4betacdot cos^2beta}{cos^4betacdot sin^2beta}=dfrac{sin^2beta}{cos^2beta}={bf tg^2beta}$