Question

Ctg a =-12/13, п/2<a<п. Найдите ​

Answer 1

решение смотри на фотографии

Answer 2

$1)tg\alpha*Ctg\alpha=1\\\\tg\alpha=\frac{1}{Ctg\alpha }=-\frac{13}{12} =-1\frac{1}{12}\\\\\boxed{tg\alpha=-1\frac{1}{12}}\\\\\\2)1+Ctg^{2}\alpha=\frac{1}{Sin^{2}\alpha}\\\\Sin^{2}\alpha=\frac{1}{1+Ctg^{2}\alpha}=\frac{1}{1+\frac{144}{169}}=\frac{1}{\frac{313}{169}} =\frac{169}{313}\\\\Sin\alpha0 \ \Rightarrow \ Sin\alpha=\sqrt{\frac{169}{313}}=\frac{13}{\sqrt{313}} \\\\\boxed{Sin\alpha=\frac{13}{\sqrt{313}}}\\\\\\3)Ctg\alpha=\frac{Cos\alpha }{Sin\alpha}$

$Cos\alpha=Ctg\alpha*Sin\alpha=-\frac{12}{13}*\frac{13}{\sqrt{313}} =-\frac{12}{\sqrt{313}}\\\\\boxed{Cos\alpha=-\frac{12}{\sqrt{313}}}$