Не могу решить 8 и 9

Решите задачу:

$8) \\ \frac{1-sin^4 \alpha -cos^4 \alpha }{cos^2 \alpha } = \frac{(1^2-(sin^2 \alpha )^2)-cos^4 \alpha }{cos^2 \alpha } = \\ \\ = \frac{(1-sin^2 \alpha )(1+sin^2 \alpha )-cos^4 \alpha }{cos^2 \alpha }= \frac{cos^2 \alpha (1+sin^2 \alpha )-cos^4 \alpha }{cos^2 \alpha }= \\ \\ = \frac{cos^2 \alpha (1+sin^2 \alpha )}{cos^2 \alpha }- \frac{cos^4 \alpha }{cos^2 \alpha }= \\ \\ =1+sin^2 \alpha -cos^2 \alpha =(1-cos^2 \alpha )+sin^2 \alpha =sin^2 \alpha +sin^2 \alpha =2sin^2 \alpha$

$9. \\ a) tg1*tg89=tg1*tg(90-1)=tg1*ctg1=tg1* \frac{1}{ctg1}=1 \\ \\ b) tg3*tg87=tg3*tg(90-3)=tg3*ctg3=tg3* \frac{1}{tg3}=1 \\ \\ tg1*tg3*tg5*...*tg89=1$