Dwa proste zadanka w załączniku. Daje wszystkie pkt i naj ;D

6.18 [latex]a)\ 1+ctg^2alpha=frac{1}{sin^2alpha}\ L=1+frac{cos^2}{sin^2}=frac{sin^2}{sin^2}+frac{cos^2}{sin^2}=frac{sin^2+cos^2}{sin^2}=frac{1}{sin^2}\ L=P\ b)\ frac{sin}{1+cos}-frac{1-cos}{sin}=0\ L=frac{sin^2}{sin(1+cos)}-frac{(1-cos)(1+cos)}{sin(1+cos)}=frac{sin^2}{sin(1+cos)}-frac{1-cos^2}{sin(1+cos)}=\ =frac{0}{sin(1+cos)}=0\ L=P\ [/latex] [latex]c)\ frac{1}{tg + ctg}=sincdot cos\ L=frac{1}{frac{sin}{cos}+frac{cos}{sin}}=frac{1}{frac{sin^2}{sin cdot cos}+frac{cos^2}{sin cdot cos}}=frac{sin cdot cos}{sin^2+cos^2}=sin cdot cos\ L=P\ d)\ frac{sin^2}{1-cos}=1+cos\ L=frac{1-cos^2}{1-cos}=frac{(1-cos)(1+cos)}{1-cos}=1+cos\ L=P [/latex]   6.19 [latex]a)\ sin^2+cos^2=1\ (frac{1}{5})^2+cos^2=1\ frac{1}{25}+cos^2=1\ cos^2=1-frac{1}{25}=frac{24}{25}\ cos=sqrt{frac{24}{25}}=frac{2sqrt{6}}{5}\ \ ctg=frac{cos}{sin}=frac{frac{2sqrt{6}}{5}}{frac{1}{5}}=2sqrt{6}\ \ frac{ctg-cos}{ctg}=frac{2sqrt{6}-frac{2sqrt{6}}{5}}{2sqrt{6}}=frac{frac{10sqrt{6}-2sqrt{6}}{5}}{2sqrt{6}}=frac{8sqrt{6}}{10sqrt{6}} = frac{4}{5}\ \ [/latex] [latex]b)\ tg=frac{sin}{cos}\ sqrt{2}=frac{sin}{cos}\ sin=sqrt{2}cos\ \ sin^2+cos^2=1\ (sqrt{2}cos)^2+cos^2=1\ 2cos^2+cos^2=1\ 3cos^2=1\ cos^2=frac{1}{3}\ cos=sqrt{frac{1}{3}}=frac{sqrt{3}}{3}\ \ frac{1}{cos}-cos=frac{3}{sqrt{3}}-frac{sqrt{3}}{3}=frac{2sqrt{3}}{3} [/latex]