Proszę o pomoc zadanie 138 139 oraz 845 od a do c

w zadaniu 139 podpunkt b) nie zmieścił mi się na kartce :) więc napisałem, że drugi nawias jest do potęgi minus pierwszej, co jest tożsame z napisaniem tego jako licznika ułamka. 

[latex] 138.\sin^{2}alpha+cos^{2}alpha = 1[/latex] [latex]a)\L = (ctg^{2}alpha+1)sin^{2}alpha = (frac{cos^{2}alpha}{sin^{2}apha}+1)sin^{2}alpha = frac{cos^{2}alpha*sin^{2}alpha}{sin^{2}alpha}+sin^{2}alpha =\\=cos^{2}alpha+sin^{2}alpha = 1 = P[/latex] [latex]b)\L = cos^{4}alpha-sin^{4}alpha = (cos^{2}alpha-sin^{2}alpha)(cos^{2}alpha+sin^{2}alpha) = \\=(cos^{2}alpha-sin^{2}alpha)*1 = cos^{2}alpha-sin^{2}alpha = P[/latex] [latex]c)\L = (cosalpha+sinalpha)^{2}+(cosalpha-sinalpha)^{2} =\\=cos^{2}alpha+2sinalphacosalpha+sin^{2}alpha+cos^{2}alpha-2cosalphasinalpha+sin^{2}alpha =\\= 2(sin^{2}alpha+cos^{2}alpha) = 2*1 = 2 = P[/latex] [latex] 139.\a)\\P = 1+ctg^{2}alpha = 1+frac{cos^{2}alpha}{sin^{2}alpha} =\\= frac{sin^{2}alpha+cos^{2}alpha}{sin^{2}alpha} = frac{1}{sin^{2}alpha} = L[/latex] [latex]b)\\L = frac{1}{1+tg^{2}alpha}+frac{1}{1+ctg^{2}alpha} =frac{1}{1+frac{sin^{2}alpha}{cos^{2}alpha}}+frac{1}{1+frac{cos^{2}alpha}{sin^{2}alpha}}=\\=frac{1}{frac{cos^{2}alpha+sin^{2}alpha}{cos^{2}alpha}}+frac{1}{frac{sin^{2}alpha+cos^{2}alpha}{sin^{2}alpha}}=frac{sin^{2}alpha+cos^{2}alpha}{sin^{2}alpha+cos^{2}alpha} = frac{1}{1} = 1 = P[/latex] [latex]845.\a)\\L = frac{sin(90^{o}-alpha)}{sinalpha} = frac{cosalpha}{sinalpha} = ctgalpha\\P = ctgalpha\\L = P[/latex] [latex]b)\\L = tg(90^{o}-alpha)tgalpha = ctgalpha*tgalpha = 1\\P = 1\\L = P[/latex] [latex]c)\\L = sinalphacos(90^{o}-alpha) = sinalpha*sinalpha= sin^{2}alpha\\P = 1-sin(90^{o}-alpha)cosalpha = 1-cosalpha*cosalpha = \=sin^{2}alpha+cos^{2}alpha-cos^{2}alpha =sin^{2}alpha\\L = P[/latex]