Ребята помогите докажите тождество 470 задача

1) 4sinα*cos³α-2sin2α*sin²α=sin4α;
4sinα*cos³α-2sin2α*sin²α=4sinα*cos³α-4sinα*cosα*sin²α=
=4sinα*cosα(cos²α-sin²α)=4sinα*cosα*cos2α=2sin2α*cos2α=sin4α;
$2) \frac{tgx}{1+ tg^{2}x}+ \frac{ctgx}{1+ctg^2x}=sin2x; \\ tgx*cos^2x+ctgx*sin^2x=sinx*cosx+sinx*cosx= \\ =2sinx*cosx=sin2x;$
3) tg^4α(8cos²(π-α)-cos(π+4α)-1)=8sin^4α;
tg^4α(8cos²α+cos4α-1)=tg^4α(8cos²α+cos²2α-sin²2α-sin²2α-cos²2α)=
=tg^4α(8cos²α-2sin²2α)=tg^4α(8cos²α-8sin²α*cos²α)=8sin^4α/cos²α(1-sin²α)=
=8sin^4α/cos²α*cos²α=8sin^4α;
$4) 2tg \frac{ \alpha }{2}(tg \alpha +ctg \alpha )(1-tg^2 \frac{ \alpha }{2})= \frac{1}{ cos^{4} \frac{ \alpha }{2}}; \\ 2tg \frac{ \alpha }{2}( \frac{2tg \frac{ \alpha }{2}}{1-tg^2 \frac{ \alpha }{2}}+ \frac{1-tg^2 \frac{ \alpha }{2} }{2tg^2 \frac{ \alpha }{2}})(1-tg^2 \frac{ \alpha }{2})= ( \frac{4tg^2 \frac{ \alpha }{2} }{1-tg^2 \frac{ \alpha }{2}}+(1-tg^2 \frac{ \alpha }{2}))(1-tg^2 \frac{ \alpha }{2})$
$=4tg^2 \frac{ \alpha }{2}+ (1-tg^2 \frac{ \alpha }{2} )^{2}=4tg^2 \frac{ \alpha }{2}+1-2tg^2 \frac{ \alpha }{2}+tg^4 \frac{ \alpha }{2}= \\ =1+2tg^2 \frac{ \alpha }{2}+tg^4 \frac{ \alpha }{2}= (1+tg^2 \frac{ \alpha }{2} )^{2}= \frac{1}{cos^4 \frac{ \alpha }{2}}.$