Question

tan A tan(60+A)tan(120+a)=-tan3A

Answer 1

TANA +TAN(60+A)+TAN(120+A)=3TANA

Answers :

LHS = tanA +[ [tanA+sqrt3] / [1-{(sqrt3)tanA)}] ] +[ [tanA-sqrt3] / [1+[sqrt3}tanA ] ]

ON solving

=> tanA + [8tanA /(1-3sqtanA)]

=> 9tanA-3cubetanA / [1-3sqtanA]

=> 3tan3A =RHS

Tan A + (tan 60 + tan A )/(1-tan60.tanA) + (tan (120) + tan A )/(1-tan120.tanA)

tan A + [root(3) + tanA]/[1-root(3).tanA] + [ tan A - root(3) ] /[1 + root(3).tanA]

tanA + [ root(3) + 3.tanA + tanA + root(3).tan^2A + tanA - root(3) -root(3).tan^2A + 3.tanA ] / [ 1 - 3tan^2A ]

tanA + [ 8tanA] / [ 1-3tan^2A ]

[9tanA - 3.tan^3A ]/[1-3tan^2A]

= 3.tan3A ( eqaul to RHS )

Hence proved