7x^2-8x+1=0
(7x-1)(x-1)=0
x1=1/7,x2=1
所以设tana=1,tanb=1/7.
或者由韦达定理得tana+tanb=8/7,tanatanb=1/7
tan(a+b)=(tana+tanb)/(1-tanatanb)=(8/7)/(1-1/7)=4/3.
tan(a+b)=2[tan(a+b)/2]/[1-(tan(a+b)/2)^2]
设tan(a+b)/2=t
即2t/(1-t^2)=4/3
3t=2-2t^2
2t^2+3t-2=0
(2t-1)(t+2)=0
t=1/2或-2.
即tan(a+b)/2=1/2或-2