tan（π +a)=-1/3,tan(a+b)=sin(π-2a)+4cos^2a/10cos^2a-sin2a.求ta

tan（π +a)=-1/3 ===>tana=-1/3
因为tan(a+b)=(sin2a+4cosa)/(10cosa-sin2a)
=(2sinacosa+4cosa)/(10cosa-2sinacosa)
=(2tana+4)/(10-2tana)
=(2×(-1/3)+4)/(10-2×(-1/3))
=5/16
因为tan(a+b)=(tana+tanb)/(1-tanatanb)
则有(-1/3+tanb)/(1+tanb/3)=5/16
然后自己算出tan