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Question 1

Question 2

1
сtga=-5/12,a-2ч
sin²a=1:(1+ctg²a)=1:(1+25/144)=1:169/144=144/169
sina=12/13
cosa=-√(1-sin²a)=-√(1-144/169)=-√(25/169)=-5/13
tga=1:ctga=1:(-5/12)-12/5=-2,4
2
(sina/cosa-cosa/sina):(sina-cosa)=(sin²a-cos²a)/(sinacosa):*1/(sina-cosa)=
=(sina-cosa)(sina+cosa)/[sinacosa(sina-cosa)]=(sina+cosa)/(sinacosa)=
=(sina+cosa):[(sina+cosa)²-1]/2=1,2:(1,2²-1)/2=1,2:(1,44-1)/2=1,2:0,22=
=120:22=5 5/11
3
(-tga-1)(ctga+1)/[tga+1)(-ctga-1)]=-(tga+1)(ctga+1)/[-(tga+1)(ctga+1)]=1

Question 3

=========== 1 ===========
$ctg \alpha =- \frac{5}{12} \rightarrow tg \alpha = \frac{1}{ctg \alpha } = - \frac{12}{5}=- 2.4\\\\ 1+tg^2 \alpha = \frac{1}{cos^2 \alpha } \rightarrow \\\\ cos^2 \alpha = \frac{1}{1+tg^2 \alpha} \\\\ \frac{ \pi }{2} \ \textless \ \alpha \ \textless \ \pi \Rightarrow cos \alpha \ \textless \ 0 \Rightarrow\\\\ cos \alpha =- \sqrt{ \frac{1}{1+tg^2 \alpha}} = - \sqrt{ \frac{1}{1+(- \frac{12}{5} )^2}} = - \sqrt{ \frac{1}{1+ \frac{144}{25} }} = - \sqrt{ \frac{1}{ \frac{169}{25} }} = \\\\ =- \sqrt{ \frac{25}{169} }=- \frac{5}{13} \\\\$
$sin^2 \alpha + cos^2 \alpha = 1\\ sin^2 \alpha = 1 - cos^2 \alpha\\ \frac{ \pi }{2} \ \textless \ \alpha \ \textless \ \pi \Rightarrow sin \alpha \ \textgreater \ 0 \Rightarrow\\ sin \alpha = \sqrt{1 - cos^2 \alpha} = \sqrt{1-(- \frac{5}{13} )^2} = \sqrt{1-\frac{25}{169}} = \sqrt{ \frac{144}{169} }= \frac{12}{13}$

sedinalana_zn · Answer 1 · 2018-05-16T02:10:04+0000

1
сtga=-5/12,a-2ч
sin²a=1:(1+ctg²a)=1:(1+25/144)=1:169/144=144/169
sina=12/13
cosa=-√(1-sin²a)=-√(1-144/169)=-√(25/169)=-5/13
tga=1:ctga=1:(-5/12)-12/5=-2,4
2
(sina/cosa-cosa/sina):(sina-cosa)=(sin²a-cos²a)/(sinacosa):*1/(sina-cosa)=
=(sina-cosa)(sina+cosa)/[sinacosa(sina-cosa)]=(sina+cosa)/(sinacosa)=
=(sina+cosa):[(sina+cosa)²-1]/2=1,2:(1,2²-1)/2=1,2:(1,44-1)/2=1,2:0,22=
=120:22=5 5/11
3
(-tga-1)(ctga+1)/[tga+1)(-ctga-1)]=-(tga+1)(ctga+1)/[-(tga+1)(ctga+1)]=1