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 \(1+\ctg^2\alpha=1/\sin^2\alpha\).

а) \(\frac{\sin 2\alpha}{\cos^2\alpha-\sin^2\alpha}=\frac{\sin 2\alpha}{\cos 2\alpha}= g 2\alpha\).
б) \(\cos 2\alpha+\sin 2\alpha\cdot g\alpha=\cos^2\alpha-\sin^2\alpha+2\sin^2\alpha=1\).
в) \(\frac{\cos 2\alpha-1}{\sin\alpha\cos\alpha+\sin^2\alpha}=\frac{-2\sin^2\alpha}{\sin\alpha(\cos\alpha+\sin\alpha)}=-\frac{2\sin\alpha}{\cos\alpha+\sin\alpha}\).
г) \(\frac{\sin 2\alpha-2\cos\alpha}{\sin\alpha-\sin^2\alpha}=\frac{2\cos\alpha(\sin\alpha-1)}{\sin\alpha(1-\sin\alpha)}=-2\ctg\alpha\).
д) \( g\alpha(1+\cos 2\alpha)=\frac{\sin\alpha}{\cos\alpha}\cdot2\cos^2\alpha=\sin 2\alpha\).
е) \(\frac{\sin\alpha+\sin 2\alpha}{1+\cos\alpha+\cos 2\alpha}=\frac{\sin\alpha(1+2\cos\alpha)}{\cos\alpha(1+2\cos\alpha)}= g\alpha\).
ж) \((1-\sin^2\alpha)(1+ g^2\alpha)=\cos^2\alpha\cdot\frac1{\cos^2\alpha}=1\).
з) \(1+\frac{\cos^4\alpha+\sin^2\alpha\cos^2\alpha}{\sin^2\alpha}=1+\frac{\cos^2\alpha(\cos^2\alpha+\sin^2\alpha)}{\sin^2\alpha}=1+\ctg^2\alpha=\frac1{\sin^2\alpha}\).
и) \(\sin^2\alpha(1+\ctg^2\alpha)-\cos^2\alpha=1-\cos^2\alpha=\sin^2\alpha\).
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) \(1/\cos^2\alpha\).

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