Odpowiedź:
twierdzenie pitagorasa : [tex]a^{2}[/tex] + [tex]b^{2}[/tex] = [tex]c^{2}[/tex]
b = 3y
c = [tex]\sqrt{10y}[/tex]
[tex]a^{2}[/tex] + [tex]3y^{2}[/tex] = [tex](\sqrt{10y}) ^{2}[/tex]
[tex]a^{2}[/tex] + 9y = 10y
[tex]a^{2}[/tex] = 10y - 9y
[tex]a^{2}[/tex] = y
a = [tex]\sqrt{y}[/tex]
ctg a = [tex]\frac{3y}{y}[/tex] = 3
[tex]sin^{2}[/tex] a = [tex](\frac{y}{\sqrt{10y}})^{2}[/tex]
[tex]cos^{2}[/tex] a = [tex](\frac{3y }{\sqrt{10y}})^{2}[/tex]
2tg a = 2[tex]\frac{y}{3y}[/tex] = [tex]\frac{2}{3}[/tex]
[tex]\frac{ctg a + sin^{2}a}{cos^{2} - 2tg a}[/tex] = [tex]\frac{3 + (\frac{y}{\sqrt{10y}})^{2}}{(\frac{3y }{\sqrt{10y}})^{2} - \frac{2}{3} }[/tex] = [tex]\frac{3 + \frac{1}{10} }{\frac{9}{10} - \frac{2}{3} }[/tex] = [tex]\frac{3\frac{1}{10} }{\frac{27}{30} - \frac{20}{30} }[/tex] = [tex]\frac{\frac{31}{10} }{\frac{7}{30} }[/tex] = [tex]\frac{31}{10} : \frac{7}{30}[/tex] = [tex]\frac{31}{10}[/tex] * [tex]\frac{30}{7}[/tex] = [tex]\frac{91}{7}[/tex] = 13
Szczegółowe wyjaśnienie:
