Odpowiedź :
[tex]a) \\tg\alpha = \sqrt2\\tg\alpha = \frac{sin\alpha}{cos\alpha}\\\sqrt2 = \frac{sin\alpha}{cos\alpha} /*cos\alpha\\\sqrt2cos\alpha = sin\alpha\\sin^2\alpha + cos^2\alpha=1\\(\sqrt2cos\alpha)^2+cos^2\alpha=1\\2cos^2\alpha+cos^2\alpha=1\\3cos^2\alpha=1 /:3\\cos^2\alpha=\frac13\\cos\alpha=\sqrt{\frac13}\\sin\alpha=\sqrt2*\sqrt{\frac13}\\sin\alpha=\sqrt{\frac23}\\ctg\alpha*tg\alpha=1\\ctg\alpha*\sqrt2=1 /:\sqrt2\\ctg\alpha=\frac1{\sqrt2}=\frac{\sqrt2}2[/tex]
[tex]b) \\tg\alpha = \frac23\\\frac{sin\alpha}{cos\alpha}=\frac23\\2cos\alpha=3sin\alpha /:3\\\frac23cos\alpha=sin\alpha\\(\frac23cos\alpha)^2+cos^2\alpha=1\\\frac49cos^2\alpha+cos^2\alpha=1\\\frac{13}9cos^2\alpha=1/*9\\13cos^2\alpha=9 /:13\\cos^2\alpha=\frac9{13}\\cos\alpha=\sqrt{\frac9{13}}\\cos\alpha=\frac{3}{\sqrt13} = \frac{3\sqrt{13}}{13}\\sin\alpha=\frac23*\frac{3\sqrt{13}}{13}=\frac{2\sqrt{13}}{13}\\\frac23*ctg\alpha=1 /*3\\2ctg\alpha=3/:2\\ctg\alpha=\frac32[/tex]
c)
[tex]sin\alpha=0,4 = \frac{4}{10} = \frac25\\sin^2\alpha+cos^2\alpha=1\\(\frac25)^2+cos^2\alpha=1\\\frac4{25}+cos^2\alpha=1/-\frac{4}{25}\\cos^2\alpha=\frac{21}{25}\\cos\alpha=\sqrt{\frac{21}{25}} = \frac{\sqrt{21}}5\\tg\alpha=\frac{sin\alpha}{cos\alpha}\\tg\alpha=\frac25*\frac5{\sqrt{21}} = \frac2{\sqrt{21}} = \frac{2\sqrt{21}}{21}\\tg\alpha*ctg\alpha=1\\\frac{2\sqrt{21}}{21}*ctg\alpha=1/*21\\2\sqrt{21}ctg\alpha=21 /:2\sqrt{21}\\[/tex]
[tex]ctg\alpha=\frac{21}{2\sqrt{21}}=\frac{21\sqrt{21}}{2*21}=\frac{\sqrt{21}}2[/tex]