若α+β=π/4,则(1+tanα)(1+tanβ)等于

问题描述:

若α+β=π/4,则(1+tanα)(1+tanβ)等于

α+β=π/4,
tan(π/4)=tan[α+β]=[tanα+tanβ]/[1-tanαtanβ]=1
所以
tanα+tanβ=1-tanαtanβ
tanα+tanβ+tanαtanβ=1
(1+tanα)(1+tanβ)
=1+tanα+tanβ+tanαtanβ
=1+1
=2