Question

помогите решить, пожалуйста !!!вот эти два примера

Answer 1

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

Answer 2

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

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