[latex]sin alpha= frac{sqrt{3}}{2} \ (frac{sqrt{3}}{2})^2+cos^2 alpha=1 \ frac{3}{4} +cos^2 alpha=1 \ cos^2 alpha=frac{1}{4} \ cos alpha=frac{1}{2} \ tg alpha=frac{sin alpha}{cos alpha}\ tg alpha=frac{sqrt{3}}{2} cdot 2=sqrt{3} \ tg alpha+frac{sin alpha}{cos alpha}=2 tg alpha =2 cdot sqrt{3}=2sqrt{3}[/latex]
Kąt alfa jest ostry [latex]sin alpha = frac{ sqrt{3} }{2} \\Obliczamy wartosc cos alpha :\sin^{2} alpha +cos^{2} alpha =1\\(frac{ sqrt{3} }{2} )^{2}+cos^{2} alpha =1\\cos^{2} alpha +frac{3}{4}=1 /-frac{3}{4}\\cos^{2} alpha =frac{1}{4}\\cos alpha =frac{1}{sqrt{2}} o cos alpha =frac{sqrt{2}}{2}\..................................................[/latex] [latex]Obliczamy wartosc tg alpha :\tgalpha=frac{sin alpha }{cos alpha } \\tg alpha =frac{frac{ sqrt{3} }{2}}{frac{ sqrt{2} }{2}}=frac{ sqrt{3} }{ ot2_{1}}*frac{ ot2_{1}}{ sqrt{2} }=frac{ sqrt{3} }{ sqrt{2} }=frac{ sqrt{3}* sqrt{2} }{ sqrt{2} * sqrt{2} }=frac{ sqrt{6} }{2}\\tg alpha alpha=frac{ sqrt{6} }{2} \..................................................[/latex] [latex]Obliczamy wartosc wyrazenia:\tg alpha +frac{sin alpha }{cos alpha } o tg alpha =frac{sinalpha}{cosalpha}\\tg alpha +frac{sin alpha }{cos alpha }=tg alpha +tg alpha =frac{ sqrt{6} }{2}+frac{ sqrt{6} }{2}=frac{ ot2_{1} sqrt{6} }{ ot2_{1}}=sqrt{6}[/latex] Wartość wyrażenia jest równa √6
