Question

tanA+tan(60° + A). tan(120° +A) = - tan3A​

Answer 1

Answer:

${\huge{\boxed{\mathcal{\orange{The proof is as follows:</p><p>Step 1:</p><p>Given Data:  </p><p>Tan A + tan(60+A)+tan(120+A)= 3.tan3A</p><p>To Prove: LHS =RHS</p><p>Step 2:</p><p>Left hand side:</p><p>Tan A +   (tan 60 + tan A)/(1-tan60.tanA) +  (tan (120) + tan A )/(1-tan120.tanA)</p><p>Step 3:</p><p>\tan A+\frac{\sqrt{3}+\tan A}{1-\sqrt{3}+\tan A}+\frac{\tan A-\sqrt{3}+1}{1+\sqrt{3}+\tan A}tanA+1−3+tanA3+tanA+1+3+tanAtanA−3+1</p><p>Step 4:</p><p>\tan A+\frac{\left.\sqrt{3}++3 \cdot \tan A+\tan A+\sqrt{3}+\tan ^{\wedge} 2 A+\tan A-\sqrt{3}+-\sqrt{3}+\tan ^{\wedge} 2 A+3 \tan A\right]}{1-3 \tan ^{\wedge} 2 A}tanA+1−3tan∧2A3++3⋅tanA+tanA+3+tan∧2A+tanA−3+−3+tan∧2A+3tanA]</p><p>Step 5:</p><p>\tan A+\frac{8 \tan A}{1-3 \tan 2 A}tanA+1−3tan2A8tanA</p><p>Step 6:</p><p>\frac{9 \tan A-3 \cdot \tan 3 A}{1-3 \tan 2 A}1−3tan2A9tanA−3⋅tan3A</p><p>Step 7:</p><p>= 3.tan3A   (Equal to RHS)</p><p>Therefore  LHS=RHS</p><p>Hence it is satisfied</p><p>}}}}}$

Answer 2

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 )