tga,tgb为方程mx^2-(2m-3)x+m-2=0的两实根
则tga+tgb=(2m-3)/m,tga*tgb=(m-2)/m
tg(a+b)=(tga+tgb)/(1-tgatgb)
=((2m-3)/m)/(1-(m-2)/m)
=(2m-3)/2=m-3/2
方程有两实根
则判别式=(2m-3)^2-4m(m-2)=-4m+9>=0
则m