Question

Помогите решить, пожалуйста !!!вот эти два примера

Answer 1

$\frac{tg ^{2}2 \alpha -ctg ^{2}2 \alpha }{4ctg 4\alpha }= \frac{ \frac{sin ^{2} 2 \alpha }{cos ^{2}2 \alpha} - \frac{cos ^{2}2 \alpha}{sin ^{2} 2 \alpha } }{4 \frac{cos4 \alpha }{sin4 \alpha } }}= \frac{((sin ^{2}2 \alpha)^{2}- (cos ^{2}2 \alpha )^{2})sin4 \alpha }{4cos ^{2}2 \alpha \cdot sin ^{2}2 \alpha cos4 \alpha }= \\ \frac{(sin^{2}2 \alpha -cos ^{2}2 \alpha )(sin^{2}2 \alpha +cos ^{2}2 \alpha )2sin2 \alpha cos2 \alpha }{4cos ^{2}2 \alpha \cdot sin ^{2}2 \alpha cos4 \alpha }=$
$=\frac{(sin^{2}2 \alpha -cos ^{2}2 \alpha )\cdot2sin2 \alpha cos2 \alpha }{4cos ^{2}2 \alpha \cdot sin ^{2}2 \alpha cos4 \alpha }=\frac{-cos4 \alpha \cdot2sin2 \alpha cos2 \alpha }{4cos ^{2}2 \alpha \cdot sin ^{2}2 \alpha cos4 \alpha }=-\frac{sin2 \alpha cos2 \alpha }{2cos ^{2}2 \alpha \cdot sin ^{2}2 \alpha }= \\ =- \frac{1}{2sin2 \alpha cos2 \alpha }=- \frac{1}{sin4 \alpha }$

$\frac{cos \alpha -cos3 \alpha }{1-cos2 \alpha } + \frac{sin \alpha -sin3 \alpha }{sin2 \alpha }= \\= \frac{-2sin (-\alpha )sin2 \alpha }{2sin ^{2} \alpha } + \frac{2\cdot sin(- \alpha )(cos2 \alpha ) }{2sin \alpha\cdot cos \alpha } = \\ =2cos \alpha -\frac{ cos2 \alpha }{ cos \alpha } = \frac{2cos ^{2} \alpha -cos2 \alpha }{cos \alpha }= \\ = \frac{2cos ^{2} \alpha -2cos^{2} \alpha+1 }{cos \alpha }= \frac{1}{cos \alpha }$

Answer 2

1)=[(sin^4(2a)-cos^4(2a)]sin4a/sin²2a*cos²2a*4cos4a=
=-2cos4a*sin4a*4/sin²4a*4cos4a=-2/sin4a
2)=2sinasin2a/2sin²a - 2sinacos2a/2sinacosa=sin2a/sina - cos2a/cosa=
=(sin2acosa-cos2asina)/sinacosa=sina/sinacosa=1/cosa