prove (tan12)(tan48)(tan54)(tan72) = 1

we have
tanθ tan(60° - θ) tan (60° + θ) = tan3θ. (proof is below)

put θ =12 to get
tan 12 tan(60 - 12) tan (60 + 12) = tan 3 * 12
=> tan 12 tan 48 tan 72 = tan 36
=> tan 12 tan 48 tan 72 = cot (90-36) = cot 54 = 1/ tan 54
=> (tan12)(tan48)(tan54)(tan72) = 1
proved
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to prove

tanθ tan(60° - θ) tan (60° + θ) = tan3θ

tan(60° - θ) tan (60° + θ)
= ((tan 60 - tan θ)/(1+ tan 60 tan θ))((tan 60 - tan θ)/(1+ tan 60 tan θ))
= (tan^2 60 - tan^2 θ)/(1- tan^2 60 tan^2 θ))
= (3 - tan θ)/(1 - 3 tan^2 θ))

so tanθ tan(60° - θ) tan (60° + θ) = (3 tan θ - tan^3 θ)/(1 - 3 tan^2 θ)) = tan 3θ

proved