Question

已知A+B=π/4，则1+tanA+tanB+tanA*tanB的值为多少？

Answer 1

A+B=π/4
tan(π/4)=tan(A+B)
1=(tanA+tanB)/(1-tanAtanB)
1-tanAtanB=tanA+tanB
tanA+tanB+tanAtanB=1
所以 1+tanA+tanB+tanA*tanB=2

Answer 2

根据tanα+tanβ=tan(α+β)(1-tanαtanβ)
得1+tanA+tanB+tanA*tanB=1+(1-tanAtanB)tan(A+B)+tanA*tanB
=1+tan(A+B)-tanA*tanB(tan(A+B)-1)
( A+B=π/4) =2