Ответ:

1) Используем равенства \(1+ g^2\alpha=\frac1{\cos^2\alpha}\) и \(1+\ctg^2\alpha=\frac1{\sin^2\alpha}\):

\(\frac{\sin^2\alpha}{1+ g^2\alpha}+\frac{\cos^2\alpha}{1+\ctg^2\alpha}+2 g^4\alpha\cos^4\alpha\)

\(=\sin^2\alpha\cos^2\alpha+\cos^2\alpha\sin^2\alpha+2\sin^4\alpha\)

\(=2\sin^2\alpha(\cos^2\alpha+\sin^2\alpha)=2\sin^2\alpha\).

2) Так как \(\sin2\alpha=2\sin\alpha\cos\alpha\), получаем:

\(\frac12\sin2\alpha g\alpha+\cos^2\alpha=\frac12\cdot2\sin\alpha\cos\alpha\cdot\frac{\sin\alpha}{\cos\alpha}+\cos^2\alpha\)

\(=\sin^2\alpha+\cos^2\alpha=1\).

Ответ: 1) \(2\sin^2\alpha\); 2) 1.