Докажите тождество: 1+(cos 4α/tg(3/4П-2α))=sin 4α

Решите задачу:

$1+\frac{cos4\alpha}{tg(\frac{3\pi}{4}-2\alpha)}=1+\frac{2cos4\alpha (1-tg^2(\frac{3\pi}{4}-2\alpha))}{2tg(\frac{3\pi}{4}-2\alpha)(1-tg^2(\frac{3\pi}{4}-2\alpha))}=1+\frac{2cos4\alpha}{tg(\frac{3\pi}{2}-4\alpha)(1-tg^2(\frac{3\pi}{4}-2\alpha))}=1+\frac{2cos4\alpha}{ctg4\alpha(1-tg^2(\frac{3\pi}{4}-2\alpha))}=1+\frac{2sin4\alpha}{1-tg^2(\frac{3\pi}{4}-2\alpha)}=1+\frac{2sin4\alpha cos^2(\frac{3\pi}{4}-2\alpha)}{cos^2(\frac{3\pi}{4}-2\alpha)-sin^2(\frac{3\pi}{4}-2\alpha)}=$ $1+\frac{2sin4\alpha cos^2(\frac{3\pi}{4}-2\alpha)}{cos2(\frac{3\pi}{4}-2\alpha)}=1+\frac{sin4\alpha (1+cos2(\frac{3\pi}{4}-2\alpha))}{cos(\frac{3\pi}{2}-4\alpha)}=1+\frac{sin4\alpha (1+cos2(\frac{3\pi}{4}-2\alpha))}{-sin(4\alpha)}=1-(1+cos(\frac{3\pi}{2}-4\alpha))=1- (1-sin4\alpha)=1- 1+sin4\alpha=sin4\alpha$