Question

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

Answer 1

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