Siemka prosze o rozwiazanie tych 3 zadań. Dzięki i za najlepsze naj :)

zadanie 1 kąt z IV ćwiartki [latex]left(frac{1+cosalpha}{sqrt{1-cos^2alpha}}-frac{sinalpha}{1+sqrt{1-sin^2alpha}} ight)cdot sinalpha=[/latex] [latex]left(frac{1+cosalpha}{sqrt{sin^2alpha}}-frac{sinalpha}{1+sqrt{cos^2alpha}} ight)cdot sinalpha=[/latex] [latex]left(frac{1+cosalpha}{|sinalpha|}-frac{sinalpha}{1+|cosalpha|} ight)cdot sinalpha=[/latex] [latex]left(frac{1+cosalpha}{-sinalpha}-frac{sinalpha}{1+cosalpha} ight)cdot sinalpha=[/latex] [latex]left(frac{-(1+cosalpha)}{sinalpha}-frac{sinalpha}{1+cosalpha} ight)cdot sinalpha=[/latex] [latex]frac{-(1+cosalpha)^2-sin^2alpha}{sinalpha(1+cosalpha)}cdot sinalpha=[/latex] [latex]frac{-(1+2cosalpha+cos^2alpha)-sin^2alpha}{1+cosalpha}=[/latex] [latex]frac{-1-2cosalpha-cos^2alpha-sin^2alpha}{1+cosalpha}=[/latex] [latex]frac{-1-2cosalpha-(cos^2alpha+sin^2alpha)}{1+cosalpha}=frac{-1-2cosalpha-1}{1+cosalpha}=[/latex] [latex]frac{-2-2cosalpha}{1+cosalpha}=frac{-2(1+cosalpha)}{1+cosalpha}=-2[/latex] ======================== Zadanie 2 a) [latex]cos 1590^ocdot tg(-840^o)=cos (4 cdot 360^o+150^o)cdot (-tg840^o)=[/latex] [latex]cos 150^ocdot (-tg(4 cdot180+ 120^o))=cos (180^o-30^o)cdot (-tg(120^o))=[/latex] [latex]cos (180^o-30^o)cdot (-tg(120^o))=-cos 30^ocdot (-tg(180^o-60^o))= - frac{ sqrt{3} }{2} cdot tg 60^o=- frac{ sqrt{3} }{2} cdot sqrt{3} =- frac{3}{2}=-1,5[/latex] b) Kąt z III ćwiartki [latex]log_{frac{sqrt3}{27}}tgalpha=frac{1}{5}[/latex] [latex]left( frac{sqrt3}{27} ight)^{frac{1}{5}}=tgalpha[/latex] [latex]tgalpha=left( frac{3^{ frac{1}{2} }}{3^3} ight) ^{ frac{1}{5} }[/latex] [latex]tgalpha=left(3^{-2 frac{1}{2} } ight) ^{ frac{1}{5} }[/latex] [latex]tgalpha=left(3^{-frac{5}{2} } ight) ^{ frac{1}{5} }[/latex] [latex]tgalpha=3^{-frac{1}{2} }[/latex] [latex]tgalpha=left( frac{1}{3} ight) ^{frac{1}{2} }[/latex] [latex]tgalpha= sqrt{frac{1}{3}}[/latex] [latex]tgalpha=frac{1}{ sqrt{3} }[/latex] [latex]tgalpha=frac{ sqrt{3} }{3}[/latex] [latex]alpha=180^o+30^o[/latex] [latex]alpha=210^o[/latex] ======================== Zadanie 3 Liczba [latex]3-frac{5}{m}[/latex] ma być naturalna, więc [latex]m[/latex] musi być całkowitym dzielnikiem [latex]5[/latex] [latex]m=-5 Rightarrow 3-frac{5}{m}=3-frac{5}{-5}=3+1=4in N[/latex] [latex]m=-1 Rightarrow 3-frac{5}{m}=3-frac{5}{-1}=3+5=8in N[/latex] [latex]m=1 Rightarrow 3-frac{5}{m}=3-frac{5}{1}=3-5=-2 ot in N[/latex] [latex]m=5 Rightarrow 3-frac{5}{m}=3-frac{5}{5}=3-1=2in N[/latex] [latex]m in left{ -5;-1;5 ight}[/latex]