Question

Помогите пожалуйста с правой частью времени вообще нету не успею вообще просто перегрузка жоская

Answer 1

$1)Sin\alpha=-\sqrt{1-Cos^{2}\alpha}=-\sqrt{1-(\frac{8}{17})^{2}}=-\sqrt{1-\frac{64}{289}}=-\sqrt{\frac{225}{289}}=-\frac{15}{17}\\\\tg\alpha =\frac{Sin\alpha}{Cos\alpha}=-\frac{15}{17}*\frac{17}{8}=-\frac{15}{8}=-1,875\\\\Ctg\alpha=\frac{1}{tg\alpha}=-\frac{8}{15}$

$2)1+tg^{2}\alpha=\frac{1}{Cos^{2}\alpha}\\\\Cos^{2}\alpha=\frac{1}{1+tg^{2}\alpha}=\frac{1}{1+(-\frac{\sqrt{3}}{3})^{2}}=\frac{1}{1+\frac{1}{3}}=\frac{3}{4}\\\\Cos\alpha=-\sqrt{\frac{3}{4}}=-\frac{\sqrt{3}}{2}\\\\Sin\alpha=\sqrt{1-Cos^{2}\alpha} =\sqrt{1-(-\frac{\sqrt{3}}{2})^{2}}=\sqrt{1-\frac{3}{4}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\\\\Ctg\alpha=\frac{1}{tg\alpha}=-\frac{3}{\sqrt{3}}=-\sqrt{3}$

$3)\frac{Cos\alpha}{1+Sin\alpha}+tg\alpha  =\frac{Cos\alpha}{1+Sin\alpha}+\frac{Sin\alpha}{Cos\alpha}=\frac{Cos^{2}\alpha+Sin\alpha+Sin^{2}\alpha}{Cos\alpha(1+Sin\alpha)}=\frac{1+Sin\alpha}{Cos\alpha(1+Sin\alpha)}=\frac{1}{Cos\alpha}$

$4)Ctg^{2}\alpha (Cos^{2}\alpha-1)+1=Ctg^{2}\alpha*(-Sin^{2}\alpha)+1=-\frac{Cos^{2}\alpha}{Sin^{2}\alpha}*Sin^{2}\alpha+1=-Cos^{2}\alpha+1=1-Cos^{2}\alpha=Sin^{2}\alpha$

$5)1-Sin\alpha Cos\alpha tg\alpha =1-Sin\alpha Cos\alpha *\frac{Sin\alpha}{Cos\alpha}=1-Sin^{2}\alpha=1-(0,7)^{2}=1-0,49=0,51$

Answer 2

Ответ и решение во вложении