PILNE!!!!!!! kto rozwiąże mi zadanie 5, potrzebuje na jutro ;o z góry dziekuje za pomoc

[latex]a)[/latex] Najpierw zauważmy, że skoro [latex]alpha in (90^{circ},180^{circ})[/latex], to: [latex]sin alpha extgreater 0,cos alpha extless 0.[/latex] Bogatsi o te informacje zajmijmy się naszym wyrażeniem: [latex]left(dfrac{cos alpha}{1+sqrt{1-cos^2alpha}}-dfrac{1-sin alpha}{sqrt{1-sin^2alpha}} ight)cdot cos alpha=\ =left(dfrac{cos alpha}{1+sqrt{sin^2alpha}}-dfrac{1-sin alpha}{sqrt{cos^2alpha}} ight)cdot cos alpha=\ =left(dfrac{cos alpha}{1+vert sinalpha vert}-dfrac{1-sin alpha}{vert cos alpha vert } ight)cdot cos alpha=\ =left(dfrac{cos alpha}{1+ sinalpha }-dfrac{1-sin alpha}{- cos alpha } ight)cdot cos alpha=[/latex] [latex]left(dfrac{cos alpha}{1+ sinalpha }+dfrac{1-sin alpha}{cos alpha } ight)cdot cos alpha=\ =left(dfrac{cos alpha cdot cosalpha}{(1+ sinalpha) cdot cosalpha}+dfrac{(1-sin alpha) cdot (1+ sinalpha)}{cos alpha cdot (1+ sinalpha) } ight)cdot cos alpha=\ =left(dfrac{cos^2 alpha }{(1+ sinalpha) cdot cosalpha}+dfrac{1-sin^2 alpha}{cos alpha cdot (1+ sinalpha) } ight)cdot cos alpha=[/latex] [latex]=left(dfrac{cos^2 alpha +(1-sin^2 alpha)}{(1+ sinalpha) cdot cosalpha} ight)cdot cos alpha=left(dfrac{cos^2 alpha +(1-sin^2 alpha)}{(1+ sinalpha) } ight)=\ =left(dfrac{cos^2 alpha +cos^2 alpha}{1+ sinalpha } ight)=oxed{dfrac{2cos^2 alpha}{1+ sinalpha }}[/latex] [latex]b)[/latex] Przypomnijmy wzory, z których za chwilę skorzystamy. Jeżeli [latex]alpha in (0,90^{circ})[/latex], to: [latex]sin(180^{circ}-alpha)=sin alpha\ cos(180^{circ}-alpha)=-cos alpha\ tg (180^{circ}-alpha)=-tg alpha[/latex] Wobec tego: [latex]dfrac{sin^2 120^{circ}+cos 150^{circ}}{3tg 135^{circ}-cos 60^{circ}}=dfrac{(sin 120^{circ})^2+cos 150^{circ}}{3tg 135^{circ}-frac{1}{2}}=\ =dfrac{(sin (180^{circ}-60^{circ}))^2+cos (180^{circ}-30^{circ})}{3tg (180^{circ}-45^{circ})-frac{1}{2}}=\ =dfrac{(sin (60^{circ}))^2-cos (30^{circ})}{-3tg (45^{circ})-frac{1}{2}}=\ =dfrac{(frac{sqrt{3}}{2})^2-frac{sqrt{3}}{2}}{-3cdot 1-frac{1}{2}}=[/latex] [latex]dfrac{(frac{sqrt{3}}{2})^2-frac{sqrt{3}}{2}}{-3-frac{1}{2}}=dfrac{frac{3}{4}-frac{2sqrt{3}}{4}}{frac{-7}{2}}=dfrac{frac{3-2sqrt{3}}{4}}{frac{-7}{2}}=dfrac{3-2sqrt{3}}{4}cdot left(dfrac{-2}{7} ight)=\ =dfrac{3-2sqrt{3}}{2}cdot left(dfrac{-1}{7} ight)=dfrac{2sqrt{3}-3}{14}=oxed{dfrac{sqrt{3}}{7}-dfrac{3}{14}}[/latex]

PILNE!!!!!!! kto rozwiąże mi zadanie 5, potrzebuje na jutro ;o z góry dziekuje za pomoc

PILNE!!!!!!! kto rozwiąże mi zadanie 5, potrzebuje na jutro ;o z góry dziekuje za pomoc...