2012/022) Prove that tanA + tan(π/3 +A) - tan(π/3 -A) =3tan3A

LHS
= tan(A) + tan(A+60°) - tan(60-A)

= tan(x) + (tan(x) + tan(60))/(1 - tan(x)tan(60)) - (tan 60 - tan(A))/(1 + tan60tan(A))

= tan(x) + (tan(x) + tan(60))/(1 - tan(x)tan(60)) + (tan A - tan 60)/(1 + tan60tan(A))

= tan(x) + (tan(x) + √3)/(1 - √3tan(x)) + (tan(x) - √3)/(1 + √3tan(x))

= tan(x) + (tan(x) - √3)/(1 + √3tan(x)) + (tan(x) + √3)/(1 - √3tan(x))

= tan(x) + [(tan(x) - √3)(1 - √3tan(x)) + (tan(x) + √3)(1 + √3tan(x))]/(1-3tan²(x))

= tan(x) + [tan(x) - √3tan²(x) - √3 + 3tan(x) + tan(x) + √3tan²(x) + √3 + 3tan(x)]/(1-3tan²(x))

= tan(x) + 8tan(x)/(1-3tan²(x))

= (tan(x) - 3tan³(x) + 8tan(x))/(1-3tan²(x))

= (9tan(x) - 3tan³(x))/(1-3tan²(x))

= 3(3tan(x) - tan³(x))/(1-3tan²(x))

= 3tan(3x)

= RHS