利用正切的和角公式：tan(a+b)=(tana+tanb)/(1-tanatanb)
tan(x+π/4)+tan(x+π/4)
=2[tan(x+π/4)]
=2[tanx+tan(π/4)]/[1-tanxtan(π/4)]
=2(tanx+1)/(1-tanx)
=2(1+tanx)/(1-tanx)
至此,即告一段落.要是继续做的话,结果如下：
2(1+tanx)/(1-tanx)
=2[(cosx+sinx)/cosx]/[(cosx-sinx)/cosx]
=2(cosx+sinx)/(cosx-sinx)
=2[(cosx+sinx)^2]/[(cosx-sinx)(cosx+sinx)]
=2(1+2sinxcosx)/[(cosx)^2-(sinx)^2]
=2(1+sin2x)/cos2x
=2/cos2x+2tan2x