prove that TANA +TAN(60+A)+TAN(120+A)=3TAN3A

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

           ON solving 

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

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

           => 3tan3A = RHS

Hello Krishna !

 

It is very simple ..

 

Just expand the function ..

 

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

 

Regards

 

Yagya

askiitians_expert

Dear Student,
Please find below the solution to your problem.

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 )

Thanks and Regards