Prove the following:
tanA + tan(60° + A) + tan(120° + A) = 3 tan 3A
Solution
L.H.S. = tanA + tan(60° + A) + tan(120° + A)
= `tan"A" + (tan60^circ + tan"A")/(1 - tan60^circ*tan"A") + (tan120^circ + tan"A")/(1 - tan120^circ*tan"A")`
= `tan"A" + (sqrt(3) + tan"A")/(1 - sqrt(3)tan"A") + (-sqrt(3) + tan"A")/(1 + sqrt(3)tan"A") ...[because tan60^circ = sqrt(3) and tan120^circ = tan(180^circ - 60^circ) = -tan60^circ = -sqrt(3)]`
= `(tan"A"(1 - 3tan^2"A") + (sqrt(3) + tan"A")(1 + sqrt(3)tan"A") + (-sqrt(3) + tan"A")(1 - sqrt(3)tan"A"))/((1 - sqrt(3)tan"A")(1 + sqrt(3)tan"A")`
= `(tan"A" - 3tan^3"A" + sqrt(3) + 3tan"A" + tan"A" + sqrt(3)tan^2"A" - sqrt(3) + 3tan"A" + tan"A" + sqrt(3)tan^2"A")/(1 - 3tan^2"A")`
= `(9tan"A" - 3tan^3"A")/(1 - 3tan^2"A")`
= `3((3tan"A" - tan^3"A")/(1 - 3tan^2"A"))`
= 3 tan 3A
= R.H.S.
