根据周期性：tan(5π/4) = tan[(5π/4) - π] = tan(π/4) = 1
∴原式 = {tan(π/4) + tan(5π/12)}/{1 - [tan(π/4)]·[tan(5π/12)]}
= tan[(π/4) + (5π/12)]
= tan(2π/3)
= -√3