用a,b代替吧,
(1) tan(a-b)
=(tana-tanb)/(1+tanatanb)
=(-2-1/3)/[1-2*(1/3)]
=(-7/3)/(1/3)
=-7
(2)tana<0,0tanb>0,0∴ a+b∈(π/2,3π/2)
tan(a+b)
=(tana+tanb)/(1-tanatanb)
=(-2+1/3)/[1+2*(1/3)]
=(-5/3)/(5/3)
=-1
∴ a+b=3π/4