设△ABC的内角A，B，C所对的边a，b，c成等比数列，则sinA+cosAtanCsinB+cosBtanC的取值范围是(5−12，5+12)(5−12，5+12)．

Question

设△ABC的内角A，B，C所对的边a，b，c成等比数列，则(sinA+cosAtanC)/(sinB+cosBtanC)的取值范围是

((√(5)−1)/(2)，(√(5)+1)/(2))

((√(5)−1)/(2)，(√(5)+1)/(2))

．

Answer 1

设三边的公比是q，三边为a，aq，aq2，原式=sinAcosC+cosAsinCsinBcosC+cosBsinC=sin(A+C)sin(B+C)=sinBsinA=ba=q∵aq+aq2＞a，①a+aq＞aq2②a+aq2＞aq，③解三个不等式可得q ＞5−120 ＜q＜5+12，综上有 5−12＜q＜5...