4. Упростить: а) sin 2α/(cos² α − sin² α); б) cos 2α + sin 2α·tg α; в) (cos 2α − 1)/(sin α·cos α + sin² α); г) (sin 2α − 2cos α)/(sin α − sin² α); д) tg α·(1 + cos 2α); е) (sin α + sin 2α)/(1 + cos α + cos 2α); ж) (1 − sin² α)(1 + tg² α); з) 1 + (cos⁴ α + sin² α·cos² α)/sin² α; и) sin² α·(1 + ctg² α) − cos² α; к) 1 + (sin⁴ α + sin² α·cos² α)/cos² α.

Используем формулы \(\sin 2\alpha=2\sin\alpha\cos\alpha\), \(\cos 2\alpha=\cos^2\alpha-\sin^2\alpha\), \(1-\sin^2\alpha=\cos^2\alpha\), \(1+ g^2\alpha=1/\cos^2\alpha\) и \(1+\ctg^2\alpha=1/\sin^2\alpha\).

1. а) \(\frac{\sin 2\alpha}{\cos^2\alpha-\sin^2\alpha}=\frac{\sin 2\alpha}{\cos 2\alpha}= g 2\alpha\).
2. б) \(\cos 2\alpha+\sin 2\alpha\cdot g\alpha=\cos^2\alpha-\sin^2\alpha+2\sin^2\alpha=1\).
3. в) \(\frac{\cos 2\alpha-1}{\sin\alpha\cos\alpha+\sin^2\alpha}=\frac{-2\sin^2\alpha}{\sin\alpha(\cos\alpha+\sin\alpha)}=-\frac{2\sin\alpha}{\cos\alpha+\sin\alpha}\).
4. г) \(\frac{\sin 2\alpha-2\cos\alpha}{\sin\alpha-\sin^2\alpha}=\frac{2\cos\alpha(\sin\alpha-1)}{\sin\alpha(1-\sin\alpha)}=-2\ctg\alpha\).
5. д) \( g\alpha(1+\cos 2\alpha)=\frac{\sin\alpha}{\cos\alpha}\cdot2\cos^2\alpha=\sin 2\alpha\).
6. е) \(\frac{\sin\alpha+\sin 2\alpha}{1+\cos\alpha+\cos 2\alpha}=\frac{\sin\alpha(1+2\cos\alpha)}{\cos\alpha(1+2\cos\alpha)}= g\alpha\).
7. ж) \((1-\sin^2\alpha)(1+ g^2\alpha)=\cos^2\alpha\cdot\frac1{\cos^2\alpha}=1\).
8. з) \(1+\frac{\cos^4\alpha+\sin^2\alpha\cos^2\alpha}{\sin^2\alpha}=1+\frac{\cos^2\alpha(\cos^2\alpha+\sin^2\alpha)}{\sin^2\alpha}=1+\ctg^2\alpha=\frac1{\sin^2\alpha}\).
9. и) \(\sin^2\alpha(1+\ctg^2\alpha)-\cos^2\alpha=1-\cos^2\alpha=\sin^2\alpha\).
10. к) \(1+\frac{\sin^4\alpha+\sin^2\alpha\cos^2\alpha}{\cos^2\alpha}=1+\frac{\sin^2\alpha(\sin^2\alpha+\cos^2\alpha)}{\cos^2\alpha}=1+ g^2\alpha=\frac1{\cos^2\alpha}\).

Ответ: а) \( g 2\alpha\); б) \(1\); в) \(-\frac{2\sin\alpha}{\cos\alpha+\sin\alpha}\); г) \(-2\ctg\alpha\); д) \(\sin 2\alpha\); е) \( g\alpha\); ж) \(1\); з) \(1/\sin^2\alpha\); и) \(\sin^2\alpha\); к) \(1/\cos^2\alpha\).