[tex]\displaystyle\bf\\\frac{\pi }{2} < \alpha < \pi \ \ \Rightarrow \ \ Sin\alpha > 0 \ ; \ Cos\alpha < 0 \ ; \ tg\alpha < 0\\\\\\Ctg\alpha =-\frac{12}{5} \\\\\\tg\alpha \cdot Ctg\alpha =1\\\\\\tg\alpha =\frac{1}{Ctg\alpha } =1:\Big(\frac{12}{5} \Big)=-1\cdot\frac{5}{12} =-\frac{5}{12} \\\\\\1+Ctg^{2} \alpha =\frac{1}{Sin^{2} \alpha }[/tex]

[tex]\displaystyle\bf\\Sin^{2} \alpha =\frac{1}{1+Ctg^{2}\alpha } =\frac{1}{1+\Big(-\dfrac{12}{5} \Big)^{2} } =\frac{1}{1+\dfrac{144}{25} } =\frac{1}{\dfrac{169}{25} } =\frac{25}{169} \\\\\\Sin\alpha =\sqrt{\frac{25}{169} } =\frac{5}{13}\\\\\\Sin^{2} \alpha +Cos^{2} \alpha =1\\\\\\Cos^{2}\alpha =1-Sin^{2}\alpha =1-\frac{25}{169} =\frac{144}{169} \\\\\\Cos\alpha =-\sqrt{\frac{144}{169} } =-\frac{12}{13}[/tex]

[tex]\displaystyle\bf\\Otvet:Sin\alpha =\frac{5}{13} \ ; \ Cos\alpha =-\frac{12}{13} \ ; \ tg\alpha =-\frac{5}{12}[/tex]

Ответ:

[tex]ctga=-\dfrac{12}{5}\\\\\dfrac{\pi }{2} < a < \pi \ \ \Rightarrow \ \ \ sina > 0\ ,\ cosa < 0\ ,\ tga < 0[/tex]

Известно тождество  [tex]1+ctg^2a=\dfrac{1}{sin^2a}[/tex]  .

[tex]sin^2a=\dfrac{1}{1+ctg^2a}=\dfrac{1}{1+\dfrac{144}{25}}=\dfrac{25}{169}\ \ ,\ \ sina=\dfrac{5}{13}[/tex]

Основное тригонометрическое тождество   [tex]sin^2a+cos^2a=1[/tex]  .

[tex]cos^2a=1-sin^2a=1-\dfrac{25}{169}=\dfrac{144}{169}\ \ ,\ \ \ cosa=-\dfrac{12}{13}\\\\\\tga=\dfrac{sina}{cosa}=-\dfrac{5}{12}[/tex]