如附圖，ABCD，CDEF，EFGH，GHIJ皆為正方形，若∠HBG＝α，∠AJB＝β，則tan ( α－β )?

tan (α－β) = (tan α - tan β) / (1 + tan α tan β) = (1/3 - 1/4) / (1 + 1/3 * 1/4) = 1/13.

tanα=HG/(3 BC)
tanα=HG/(3 HG)
tanα=1/3

tanβ=AB/(4 BC)
tanβ=1/4

tan(α-β)
=(tanα-tanβ)/(1+tanβtanα)
=(1/3 -1/4)/[1+(1/4)(1/3)]
=(1/12)/(13/12)
=1/13