Answer :
[tex]\alpha;\ \beta\in(0^o;\ 90^o)\\\\sin(\alpha+\beta)=1\to sin(\alpha+\beta)=sin90^o\to\alpha+\beta=90^o\\\\sin(\alpha-\beta)=\frac{1}{2}\to sin(\alpha-\beta)=sin30^o\to\alpha-\beta=30^o\\\\ +\left\{\begin{array}{ccc}\alpha+\beta=90^o\\\alpha-\beta=30^o\end{array}\right\\---------\\.\ \ \ \ \ \ \ 2\alpha=120^o\ \ \ /:2\\.\ \ \ \ \ \ \ \ \ \alpha=60^o\\\\60^o+\beta=90^o\ \ \ /-60^o\\\beta=90^o-60^o\\\beta=30^o[/tex]
[tex]tan(\alpha+2\beta)\cdot tan(2\alpha+\beta)\\\\=tan(60^o+2\cdot30^o)\cdot tan(2\cdot60^o+30^o)\\\\=tan120^o\cdot tan150^o=tan(180^o-60^o)\cdot tan(180^o-30^o)\\\\=-tan60^0\cdot(-tan30^o)=-\sqrt3\cdot(-\frac{\sqrt3}{3})=\frac{3}{3}=1\\\\\\Answer:A[/tex]
[tex]tan(\alpha+2\beta)\cdot tan(2\alpha+\beta)\\\\=tan(60^o+2\cdot30^o)\cdot tan(2\cdot60^o+30^o)\\\\=tan120^o\cdot tan150^o=tan(180^o-60^o)\cdot tan(180^o-30^o)\\\\=-tan60^0\cdot(-tan30^o)=-\sqrt3\cdot(-\frac{\sqrt3}{3})=\frac{3}{3}=1\\\\\\Answer:A[/tex]
