1.

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

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ANSWER:

{\huge{\boxed{\mathcal{\orange{The proof is as follows:Step 1:Given Data: Tan A + tan(60+A)+tan(120+A)= 3.tan3ATo Prove: LHS =RHSStep 2:Left hand side:Tan A + (tan 60 + tan A)/(1-tan60.tanA) + (tan (120) + tan A )/(1-tan120.tanA)Step 3:\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+1Step 4:\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]Step 5:\tan A+\frac{8 \tan A}{1-3 \tan 2 A}tanA+1−3tan2A8tanAStep 6:\frac{9 \tan A-3 \cdot \tan 3 A}{1-3 \tan 2 A}1−3tan2A9tanA−3⋅tan3AStep 7:= 3.tan3A (Equal to RHS)Therefore LHS=RHSHence it is satisfied}}}}}


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