tan(β-π/4)=1/4
=[tanβ-tanπ/4]/[1＋tanβtanπ/4]
=[tanβ-1]/[1＋tanβ]
4tanβ-4=1＋tanβ
3tanβ=5
tanβ=5/3
所以
tan（α+2β)
=tan【（α+β)＋β】
=【tan（α+β)＋tanβ】/[1-tan（α+β)tanβ]
=[2/5＋5/3]/[1-2/5×5/3]
=31/5