Zrobiłem zdjęcie zadania telefonem, nie rozumiem zadania kompletnie proszę o jak najszybsze udzielenie odpowiedzi

α - kąt ostry (wartości funkcji są dodatnie) [latex]sinalpha=frac{sqrt{3}}{10}\ \ \ 1^{o} cosalpha:\ \ sin^{2}alpha+cos^{2}alpha=1\ \ (frac{sqrt{3}}{10})^{2}+cos^{2}alpha-1=0\ \ cos^{2}alpha-1+frac{3}{100}=0\ \ cos^{2}alpha-frac{97}{100}=0\ \ (cosalpha-frac{sqrt{97}}{10})(cosalpha+frac{sqrt{97}}{10})=0\ \ cosalpha=frac{sqrt{97}}{10} vee cosalpha=-frac{sqrt{97}}{10} otin\ [/latex] [latex]2^{o} tgalpha:\ \ tgalpha=frac{sinalpha}{cosalpha}=frac{sqrt{3}}{10}:frac{sqrt{97}}{10}=frac{sqrt{3}}{10}*frac{10}{sqrt{97}}=frac{sqrt{3}}{sqrt{97}}*frac{sqrt{97}}{sqrt{97}}=frac{sqrt{97*3}}{97}=frac{sqrt{291}}{97}\[/latex] [latex]3^{o} ctgalpha:\ \ ctgalpha=frac{cosalpha}{sinalpha}=frac{sqrt{97}}{10}:frac{sqrt{3}}{10}= frac{sqrt{97}}{10}*frac{10}{sqrt{3}}=frac{sqrt{97}}{sqrt{3}}*frac{sqrt{3}}{sqrt{3}}=frac{sqrt{291}}{3}[/latex]

[latex]sin alpha = frac{ sqrt{3} }{10} \ \ sin^{2} alpha +cos ^{2} alpha =1 \ (frac{ sqrt{3} }{10} )^{2} +cos ^{2} alpha =1 \ cos alpha ^{2} =1-frac{ 3 }{100} \ cos alpha ^{2} =frac{97 }{100} \ cos alpha=frac{ sqrt{97 }}{10} \ \ [/latex] [latex]tg alpha = frac{sin alpha }{cos alpha} \ tg alpha =frac{frac{ sqrt{3} }{10} }{frac{ sqrt{97 }}{10} } ={frac{ sqrt{3} }{10} * frac{10}{ sqrt{97} } = frac{sqrt{3} }{sqrt{97} } * frac{sqrt{97} }{sqrt{97} }= \ = frac{ sqrt{291} }{97} [/latex] [latex]ctg alpha = frac{1}{tg alpha } =frac{ 97}{sqrt{291}}*frac{sqrt{291} }{sqrt{291}}=frac{sqrt{291} }{3}[/latex]