tanA+tanB+tanC
=tan(A+B)(1-tanAtanB)+tanC
=tan(π-c)(1-tanAtanB)+tanC
=-tanC(1-tanAtanB)+tanC
=-tanC+tanAtanBtanC+tanC
=tanAtanBtanC
2.题目有误
(sin9°+cos15°sin6°)/(cos9°-sin15°sin6°)
=(sin(15-6)+cos15*sin6)/(cos(15-6)-sin15*sin6)
=(sin15*cos6-cos15*sin6+cos15*sin6)/(cos15*cos6+sin15*sin6- sin15*sin6)
=sin15*cos6/(cos15*cos6)
=tan15 [tan(A/2)=(1-cosA)/sinA]
=(1-cos30)/sin30
=(1-√3/2)/(1/2)
=2*(1-√3/2)
=2-√3