At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Get immediate and reliable solutions to your questions from a knowledgeable community of professionals on our platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
[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]
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
