Докажите тождество tg α + ctg α + tg 3α+ ctg 3α= 8cos²2α/ sin 6α

Решите задачу:

$tgx+ctgx+tg3x+ctg3x= \frac{1-cos2x}{sin2x} +\frac{sin2x}{1-cos2x} +\frac{1-cos6x}{sin6x} +\frac{sin6x}{1-cos6x}= \\ = \frac{1-2cos2x+cos^22x+sin^22x}{(1-cos2x)sin2x}+ \frac{1-2cos6x+cos^26x+sin^26x}{(1-cos6x)sin6x}= \frac{2(1-cos2x)}{(1-cos2x)sin2x} + \\ + \frac{2(1-cos6x)}{(1-cos6x)sin6x}= \frac{2}{sin2x}+ \frac{2}{sin6x}= \frac{2(2sin4xcos2x)}{sin2xsin6x}= \frac{4 \cdot 2sin2xcos2xcos2x}{sin2xsin6x} = \\ = \frac{8cos^22x}{sin6x}$