Question

if  sin (α+β) = 1,  sin (α-β) = 1/2 then tan (α+2β) tan(2α+β) =

a. 1          b.-1            c.0              d. none

Answer 1

`\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)=(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`



`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(-(\sqrt3)/(3))=(3)/(3)=1\\\\\\Answer:A`