Oblicz wartość wyrażenia : 1. [latex]sin alpha + cos alpha sqrt{1+ tg^{2} alpha } [/latex] , jesli [latex]cos alpha= frac{5}{13} [/latex] i [latex] alpha [/latex] należy ( 0; 90 ) 2. [latex] frac{sin alpha + cos alpha }{2 sin alpha + 3 cos alpha }

1 [latex]sin^2 alpha +cos^2 alpha =1\ sin^2 alpha +(frac{5}{13})^2=1\ sin^2 alpha +frac{25}{169}=1\ sin^2 alpha =frac{144}{169}\ sin alpha =sqrt{frac{144}{169}}=frac{12}{13}\ \ tg alpha = frac{sin alpha }{cos alpha }=frac{frac{12}{13}}{frac{5}{13}} = frac{12}{5}\ \ [/latex] [latex]1\sin alpha + cos alpha sqrt{1+ tg^{2} alpha } =frac{12}{13}+frac{5}{13}sqrt{1+(frac{12}{5})^2}=frac{12}{13}+frac{5}{13}sqrt{1+frac{144}{25}}= \= frac{12}{13}+frac{5}{13}sqrt{frac{169}{25}}= frac{12}{13}+frac{5}{13}cdot frac{13}{5}=frac{12}{13}+1 = frac{25}{13}[/latex] [latex]2\ tg alpha = frac{ sqrt{2} }{2} \ frac{sin alpha }{cos alpha }=frac{sqrt{2}}{2}\ sin alpha =frac{sqrt{2}}{2}cos alpha \ sin^2 alpha +cos^2 alpha =1\ (frac{sqrt{2}}{2}cos alpha )^2+cos^2 alpha =1\ frac{1}{2}cos^2 alpha +cos^2 alpha =1\ frac{3}{2}cos^2 alpha =1\ cos^2 alpha =frac{2}{3}\ [/latex] [latex]cos alpha =sqrt{frac{2}{3}}=frac{sqrt{6}}{3}\ \ sin alpha =frac{sqrt{2}}{2}cdot frac{sqrt{6}}{3}=frac{sqrt{3}}{3}\ \ \ [/latex] [latex]frac{sin alpha + cos alpha }{2 sin alpha + 3 cos alpha } =frac{frac{sqrt{3}}{3}+frac{sqrt{6}}{3}}{2cdot frac{sqrt{3}}{3}+3cdot frac{sqrt{6}}{3}}=frac{frac{sqrt{3}+sqrt{6}}{3}}{frac{2sqrt{3}+3sqrt{6}}{3}}=frac{sqrt{3}+sqrt{6}}{3}cdot frac{3}{2sqrt{3}+3sqrt{6}}=\=frac{sqrt{3}+sqrt{6}}{2sqrt{3}+3sqrt{6}}cdot frac{2sqrt{3}-3sqrt{6}}{2sqrt{3}-3sqrt{6}} = frac{6-3sqrt{18}+2sqrt{18}-18}{12-54}=frac{-12-sqrt{18}}{-42}= [/latex] [latex]\=frac{-(12+3sqrt{2})}{-42} = frac{3(4+sqrt{2})}{42}=frac{4+sqrt{2}}{14}[/latex]