已知sina+3cosa=2,则sina-cosa/sina+cosa=
问题描述：
已知sina+3cosa=2,则sina-cosa/sina+cosa=
答
sinA+3cosA=2 --->2sin(A/2)cos(A/2)+3[cos(A/2)^2-3(sin(A/2)]^2=2[cos(A/20]^2+[sin(A/2)]^2 --->4[sin(A/2)]^2+2sin(A/2)cos(A/)-2[cos(A/2)]^2=0【两边同除2[cos(A/2)]^2】 --->2[tan(A/2)]^2+tan(A/2)-1=0 --->tan(A/2)=-1或者-1/2.tan(A/2)=-1时,sinA=2tan(A/2)/{1+[tan(A/2]^2}=-1,cosA=0 --->(sinA-cosA)/(sinA+cosA)=(-1-0)/(-1+0)=1 tan(A/2)=1/2时,tanA=2tan(A/2)/{1-[tan(A/2)]^2}=4/3 --->(sinA-cosA)/(sinA+cosA) =(tanA-1)(tanA+1) =(4/3-1)/(4/3+1) =1/7.