Prove that: tanα tan(60° - α) tan(60° + α) = tan3α

Prove that

tan α tan (60° - α) tan (60° + α) = tan3α

L.H.S. = tan α tan (60° - α) tan (60° + α)

= tan α ^{(tan 60° - tan α)}/_{(1 + tan 60° tan α)} ^{(tan 60° + tan α)}/_{(1 - tan 60° tan α)}

= tan α ^{(√3 - tan α)}/_{(1 + √3 tan α)} ^{(√3 + tan α)}/_{(1 - √3 tan α)}

= tan α ^{(3 - tan2α)}/_{(1 - 3 tan²α)}

= ^{(3 tan α - tan3α)}/_{(1 - 3 tan²α)}= tan 3α R.H.S.