Question

1)(sin2^альфа)/(1-sin^2альфа)*ctg^2альфа=
2)(sin альфа-cos альфа)^2+(cos альфа+sin альфа)^2=2

Answer 1

$\sf\displaystyle \frac{\sin^2\alpha}{1-\sin^2\alpha}\cdot ctg^2\alpha=\frac{\sin^2\alpha}{\cos^2\alpha}\cdot ctg^2\alpha=tg^2\alpha\cdot ctg^2\alpha=1$

$\tt\displaystyle (\sin\alpha-\cos\alpha)^2+(\cos\alpha+\sin\alpha)^2=2\\ \sin^2\alpha+\cos^2\alpha-2\cos\alpha\sin\alpha+\cos^2\alpha+\sin^2\alpha+2\cos\alpha\sin\alpha=2\\2cos^2\alpha+2\sin^2\alpha=2\\2=2$