Question

Алгебра. Тригонометрическая функция.

Answer 1

α - угол второй четверти, значит Sinα > 0 , Cosα < 0 , tgα < 0.

$1)tg\alpha=\frac{1}{Ctg\alpha }=\frac{1}{-\sqrt{2} }=-\frac{\sqrt{2} }{2}\\\\1+Ctg^{2}\alpha=\frac{1}{Sin^{2} \alpha }\\\\Sin^{2}\alpha=\frac{1}{1+Ctg^{2}\alpha}=\frac{1}{1+(-\sqrt{2} )^{2} }=\frac{1}{1+2}=\frac{1}{3}\\\\Sin\alpha=\sqrt{\frac{1}{3} }=\frac{1}{\sqrt{3} }=\frac{\sqrt{3} }{3}\\\\Cos\alpha=-\sqrt{1-Sin^{2}\alpha}=-\sqrt{1-\frac{1}{3} }=-\sqrt{\frac{2}{3} }=-\frac{\sqrt{6} }{3}$

$2)Sin^{2}2\alpha+Cos^{2}2\alpha+Ctg^{2}5\alpha=1+Ctg^{2}5\alpha=\frac{1}{Sin^{2}5\alpha}\\\\\frac{Ctg\alpha }{tg\alpha+Ctg\alpha}=\frac{Ctg\alpha }{\frac{1}{Ctg\alpha }+Ctg\alpha}=\frac{Ctg\alpha }{\frac{1+Ctg^{2}\alpha}{Ctg\alpha}}=\frac{Ctg^{2}\alpha}{1+Ctg^{2}\alpha}$=$\frac{\frac{Cos^{2}\alpha}{Sin^{2}\alpha} }{\frac{1}{Sin^{2}\alpha} }=\frac{Cos^{2}\alpha*Sin^{2}\alpha}{Sin^{2}\alpha}=Cos^{2}\alpha$