Question

Алгебра. Тригонометрическая функция.

Answer 1

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

$1)tgalpha=frac{1}{Ctgalpha }=frac{1}{-sqrt{2} }=-frac{sqrt{2} }{2}\\1+Ctg^{2}alpha=frac{1}{Sin^{2} alpha }\\Sin^{2}alpha=frac{1}{1+Ctg^{2}alpha}=frac{1}{1+(-sqrt{2} )^{2} }=frac{1}{1+2}=frac{1}{3}\\Sinalpha=sqrt{frac{1}{3} }=frac{1}{sqrt{3} }=frac{sqrt{3} }{3}\\Cosalpha=-sqrt{1-Sin^{2}alpha}=-sqrt{1-frac{1}{3} }=-sqrt{frac{2}{3} }=-frac{sqrt{6} }{3}$

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