Question

Помогите Решить, Дам 30 балов!

Answer 1

$2); ; dfrac{1}{cosa}-sinacdot tga=dfrac{1}{cosa}-sinacdot dfrac{sina}{cosa}=dfrac{1-sin^2a}{cosa}=dfrac{cos^2a}{cosa}=cosa\\\3); ; dfrac{sin^2a-1}{cos^2a-1}+tgacdot ctga=dfrac{-cos^2a}{-sin^2a}+1=ctg^2a+1=dfrac{1}{sin^2a}$

$4); ; cosa=-dfrac{1}{5}; ; ,; ; ain (180^circ ;270^circ ); ; Rightarrow ; ; tga>0\\1+tg^2a=dfrac{1}{cos^2a}; ; to ; ; ; 1+tg^2a=dfrac{1}{1/25}=25; ,; ; tg^2a=24\\tga=sqrt{24}=2sqrt6>0\\\5a); ; cos105+cos75=2cdot cos90cdot cos15=0\\b); ; sindfrac{11}{12}pi +sindfrac{5}{12}pi =2cdot sindfrac{4pi }{3}cdot cosdfrac{pi}{4}=2cdot (-dfrac{sqrt3}{2})cdot dfrac{sqrt2}{2}=-dfrac{sqrt6}{2}$

$c); ; cosdfrac{11}{12}pi -cosdfrac{5}{12}pi =-2cdot sindfrac{4pi }{3}cdot sindfrac{pi}{4}=2cdot dfrac{sqrt3}{2}cdot dfrac{sqrt2}{2}=dfrac{sqrt6}{2}\\\d); ; cos15-sin15=cos15-cos(90-15)=cos15-cos75=\\=-2cdot sin45cdot sin(-60)=2cdot dfrac{sqrt2}{2}cdot dfrac{1}{2}=dfrac{sqrt2}{2}$

$e); ; tg22^circ 30'+tg67^circ 30'=dfrac{sin90^circ }{cos22^circ 30'cdot cos67^circ 30'}=dfrac{1}{0,5cdot (cos90^circ +cos45^circ )}=\\\=dfrac{2}{0+frac{sqrt2}{2}}=dfrac{4}{sqrt2}=dfrac{4sqrt2}{2}=2sqrt2$

$f); ; tgfrac{11}{12}pi -tgfrac{5}{12}pi =dfrac{sinfrac{pi }{2}}{cosfrac{11pi }{12}cdot cosfrac{5pi}{12}}=dfrac{1}{0,5cdot (cosfrac{8pi }{3}+cosfrac{pi}{2})}=dfrac{2}{-frac{1}{2}+0}=-4$