Question

Помогите Пожалуйста!!!

Answer 1

$1.\a);frac{tgfracpi3}{ctgfracpi3}-sqrt2cosfrac{3pi}4=frac{frac1{sqrt3}}{frac1{sqrt3}}-sqrt2cdotleft(-frac1{sqrt2}right)=1+1=2\b);frac{sin50^o+sin10^o}{cos25^ocos5^o+sin25^osin5^o}=frac{2sinfrac{50^o+10^o}2cosfrac{50^o-10^o}2}{cos(25^o-5^0)}=frac{2sin30^ocos20^o}{cos20^o}=2cdotfrac12=1$

$2.;cosleft(frac{3pi}2+alpharight)=sinalpha=0,5\fracpi2<alpha<piRightarrowalpha=frac{5pi}6\sinleft(60^o+frac{5pi}6right)=sinleft(fracpi3+frac{5pi}6right)=sinfrac{7pi}6=-frac12$

$3.\a);left(frac{sinalpha}{tgalpha}right)^2+left(frac{cosalpha}{ctgalpha}right)^2-2sin^2alpha=left(frac{sinalpha}{frac{sinalpha}{cosalpha}}right)^2+left(frac{cosalpha}{frac{sinalpha}{cosalpha}}right)^2-2sin^2alpha=\=cos^2alpha+sin^2alpha-2sin^2alpha=cos^2alpha-sin^2alpha=cos2alpha$

$b);frac{sinalpha-sin3alpha}{cosalpha-cos3alpha}cdot(1-cos4alpha)=frac{2sinfrac{alpha-3alpha}2cosfrac{alpha+3alpha}2}{-2sinfrac{alpha+3alpha}2sinfrac{alpha-3alpha}2}cdot(1-(1-2sin^22alpha))=\=frac{2sin(-alpha)cos2alpha}{-2sin2alphasin(-alpha)}cdot(1-1+2sin^22alpha)=-frac{cos2alpha}{sin2alpha}cdot2sin^22alpha=-2sin2alphacos2alpha=\=sin4alpha$

$5.;tg2alphacdotfrac{1-tg^2alpha}{1+tg^2alpha}=tg2alphacdotcos2alpha=frac{sin2alpha}{cos2alpha}cdotcos2alpha=sin2alpha$