如果tan(a+b)=2/5,tan(b+pai/4)=1/4,求tan(a+pai/4)的值

用公式：tan(A+B)=[tan(A)+tan(B)]/[1-tan(A)tan(B)]；tan(A-B)=[tan(A)-tan(B)]/[1+tan(A)tan(B)]
设A=a+b,B=b+π/4
则 tan（A-B）=tan（a-π/4）=（2/5-1/4）/（1+2/5 * 1/4）=（3/20）/（22/20）=3/22
所以tan（a+π/4）=tan[(a-π/4)+(π/2)]=-tan[π-【(a-π/4)+(π/2)】]=-tan[π/2-（a-π/4）]=-ctg（a-π/4）=-22/3