Odpowiedź :
[tex]\alpha, \beta \in (90; 180)[/tex]
[tex]sin\alpha = \frac23\\tg\beta = -\frac23[/tex]
[tex]sin^2\alpha + cos^2\alpha = 1\\(\frac23)^2+cos^2\alpha=1\\\frac49+cos^2\alpha=1\\cos^2\alpha=1-\frac49\\cos^2\alpha=\frac59\\cos\alpha=\sqrt{\frac59}\\cos\alpha=\frac{\sqrt5}{3}\\tg\alpha = \frac{\sin\alpha}{cos\alpha}\\tg\alpha = \frac23:\frac{\sqrt5}3\\tg\alpha=\frac23*\frac3{\sqrt5}\\tg\alpha=\frac{2}{\sqrt5}=\frac{2\sqrt5}5[/tex]
[tex]tg\alpha * ctg\alpha = 1\\\frac{2\sqrt5}5*ctg\alpha=1 /*5\\2\sqrt5ctg\alpha=5 /:(2\sqrt5)\\ctg\alpha=\frac5{2\sqrt5} = \frac{5\sqrt5}{2*5} = \frac{5\sqrt5}{10}=\frac{\sqrt5}{2}[/tex]
[tex]tg\beta = \frac{sin\beta}{cos\beta}\\\frac{-2}3=\frac{sin\beta}{cos\beta}\\-2cos\beta=3sin\beta\\-\frac23cos\beta=sin\beta\\\\sin^2\beta+cos^2\beta=1\\(-\frac23cos\beta)^2+cos^2\beta=1\\\frac{4}{9}cos^2\beta+cos^2\beta=1\\\frac{13}9cos^2\beta=1 /*9\\13cos^2\beta=9 /:13\\cos^2\beta=\frac9{13}\\cos\beta=\sqrt{\frac9{13}}\\cos\beta=\frac{3}{\sqrt{13}} = \frac{3\sqrt{13}}{13}\\sin\beta = -\frac23*\frac{3\sqrt{13}}{13}\\sin\beta=-\frac{2\sqrt{13}}{13}\\[/tex]
[tex]tg\beta * ctg\beta = 1\\-\frac23*ctg\beta=1 /*3\\-2ctg\beta=3 /:(-2)\\ctg\beta = -\frac{3}2[/tex]