Ответ:
![ctg(\alpha)=-\frac{7}{24}\\sin(\alpha)=-\frac{7}{25}\\cos( \alpha)=\frac{24}{25}](https://tex.z-dn.net/?f=ctg%28%5Calpha%29%3D-%5Cfrac%7B7%7D%7B24%7D%5C%5Csin%28%5Calpha%29%3D-%5Cfrac%7B7%7D%7B25%7D%5C%5Ccos%28+%5Calpha%29%3D%5Cfrac%7B24%7D%7B25%7D)
Объяснение:
(IV четверть, ctg, tg и sin отрицательные, cos положительный.
![sin^{2}( \alpha )+cos^{2}( \alpha )=1/cos^{2}( \alpha )\\tg^{2}( \alpha)+1=\frac{1}{cos^{2}( \alpha )} \\cos^{2}( \alpha)=\frac{1}{tg^{2}( \alpha)+1} =\frac{1}{\frac{49}{576}+1 } =\frac{576}{625} \\cos( \alpha)=\frac{24}{25} \\sin(\alpha)=-\frac{7}{25} \\ctg(\alpha)=-\frac{7}{24}](https://tex.z-dn.net/?f=sin%5E%7B2%7D%28+%5Calpha+%29%2Bcos%5E%7B2%7D%28+%5Calpha+%29%3D1%2Fcos%5E%7B2%7D%28+%5Calpha+%29%5C%5Ctg%5E%7B2%7D%28+%5Calpha%29%2B1%3D%5Cfrac%7B1%7D%7Bcos%5E%7B2%7D%28+%5Calpha+%29%7D+%5C%5Ccos%5E%7B2%7D%28+%5Calpha%29%3D%5Cfrac%7B1%7D%7Btg%5E%7B2%7D%28+%5Calpha%29%2B1%7D+%3D%5Cfrac%7B1%7D%7B%5Cfrac%7B49%7D%7B576%7D%2B1+%7D+%3D%5Cfrac%7B576%7D%7B625%7D+%5C%5Ccos%28+%5Calpha%29%3D%5Cfrac%7B24%7D%7B25%7D+%5C%5Csin%28%5Calpha%29%3D-%5Cfrac%7B7%7D%7B25%7D+%5C%5Cctg%28%5Calpha%29%3D-%5Cfrac%7B7%7D%7B24%7D)
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(ах+bx)-(ay+by)=x(a+b)-y(a+b)=(a+b)(x-y)
(ax-bx)+(ay-by)=x(a-b)+y(a-b)=(a-b)(x+y)
-(ax-bx)+(ay-by)=-x(a-b)+y(a-b)=(a-b)(-x+y)
(ax-bx)-(ay-by)=x(a-b)-y(a-b)=(a-b)(x-y)
4/(a+b)+5/(a-b)-10b/(a-b)(a+b)=(4a-4b+5a+5b-10b)/(a²-b²)=
=(9a-9b)/(a²-b²)=9(a-b)/(a-b)(a+b)=9/(a=b)