СРОЧНО!!!!!Помогите упростить

Решите задачу:

$5)\frac{tg^{2}\alpha}{1+tg^{2} \alpha }*\frac{1+Ctg^{2}\alpha}{Ctg^{2}\alpha} =\frac{tg^{2}\alpha}{\frac{1}{Cos^{2}\alpha} }*\frac{\frac{1}{Sin^{2}\alpha} }{Ctg^{2}\alpha}=\frac{Sin^{2}\alpha *Cos^{2}\alpha}{Cos^{2}\alpha }*\frac{Sin^{2}\alpha}{Sin^{2} \alpha*Cos^{2} \alpha}=Sin^{2}\alpha*\frac{1}{Cos^{2} \alpha }=\frac{Sin^{2}\alpha}{Cos^{2}\alpha}=tg^{2}\alpha$

$6)\frac{1+tg\alpha }{1+Ctg\alpha }=\frac{1+\frac{Sin\alpha }{Cos\alpha } }{1+\frac{Cos\alpha }{Sin\alpha } }=\frac{\frac{Cos\alpha+Sin\alpha }{Cos\alpha } }{\frac{Cos\alpha+Sin\alpha }{Sin\alpha } }=\frac{(Cos\alpha+Sin\alpha )*Sin\alpha }{(Cos\alpha+Sin\alpha)*Cos\alpha} =\frac{Sin\alpha }{Cos\alpha } =tg\alpha$

$7)Cos^{4}\alpha+Sin^{2}\alpha Cos^{2} \alpha-Cos^{2}\alpha -1=(Cos^{4}\alpha+Sin^{2}\alpha Cos^{2}\alpha)-Cos^{2}\alpha-1=Cos^{2}\alpha(Cos^{2} \alpha+Sin^{2}\alpha)-Cos^{2}\alpha-1=Cos^{2}\alpha*1-Cos^{2} \alpha -1=Cos^{2}\alpha-Cos^{2}\alpha -1=-1$