Помогите пож 423(3,5,2,4,6,8)

3) $\frac{ctg(- \beta )\sin\beta}{\cos\beta} = -ctg \beta \cdot tg\beta=-1$
5) $ctg \alpha \sin(- \alpha )-\cos(- \alpha )=- \frac{\cos \alpha }{\sin \alpha }\cdot \sin \alpha -\cos \alpha =-2\cos \alpha$
2) $\cos^2 \alpha tg^2(- \alpha )-1=\cos^2 \alpha \cdot \frac{\sin^2 \alpha}{\cos^2 \alpha} -1=\sin^2 \alpha-1=-\cos^2 \alpha$
4) $\frac{1-tg(-x)}{\sin x+\cos(-x)} = \frac{1+tgx}{\sin x+\cos x} = \cfrac{1+ \frac{\sin x}{\cos x} }{\sin x+\cos x} = \frac{\cos x+\sin x}{\cos x(\sin x+\cos x)} = \frac{1}{\cos x}$
6) $tg(-u)ctgu+\sin^2u=-tgu\cdot ctgu+\sin^2u=-1+\sin^2u=-\cos^2u$
8) $\frac{tg(-x)+1}{1-ctgx} = \frac{-tgx+1}{1-ctgx} =\frac{tgx-1}{ctgx-1} = \cfrac{ \frac{\sin x}{\cos x} -1}{ \frac{\cos x}{\sin x}-1} = \cfrac{ \frac{\sin x-\cos x}{\cos x} }{ \frac{\cos x-\sin x}{\sin x}} =- \frac{\sin x}{\cos x} =-tgx$