1-cosx+sianx/1+cosx+sianx=-2求tanx

(1-cosx+sinx)/(1+cosx+sinx)=-2
=> cosx=-3*(sinx+1) 且 (1+cosx+sinx)!=0 (1)
cosx^2+sinx^2=9*(sinx+1)^2+sinx^2=1
=> 5*sinx^2+9*sinx+4=0
=> sinx=-1或sinx=-4/5.
若sinx=-1,则cosx=0,(1)式不成立,舍去.
若sinx=-4/5,cosx=-3*(sinx+1)=-3/5
tanx=4/3.
或
(1-cosx+sinx)/(1+cosx+sinx)
= [2sin²(x/2)+2sin(x/2)cos(x/2)]/[2cos²(x/2)+2sin(x/2)cos(x/2)]
= sin(x/2)/cos(x/2) = tan(x/2) = -2
--->tanx = 2(-2)/[1-(-2)²] = 4/3