Question

Prove that: tan A+tan(60+A)+tan(120+A)=3tan3A

Answer 1

The proof is as follows:

Step 1:

Given Data:  

Tan A + tan(60+A)+tan(120+A)= 3.tan3A

To Prove: LHS =RHS

Step 2:

Left hand side:

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

Step 3:

$\tan A+\frac{\sqrt{3}+\tan A}{1-\sqrt{3}+\tan A}+\frac{\tan A-\sqrt{3}+1}{1+\sqrt{3}+\tan A}$

Step 4:

$\tan A+\frac{\left.\sqrt{3}++3 \cdot \tan A+\tan A+\sqrt{3}+\tan ^{\wedge} 2 A+\tan A-\sqrt{3}+-\sqrt{3}+\tan ^{\wedge} 2 A+3 \tan A\right]}{1-3 \tan ^{\wedge} 2 A}$

Step 5:

$\tan A+\frac{8 \tan A}{1-3 \tan 2 A}$

Step 6:

$\frac{9 \tan A-3 \cdot \tan 3 A}{1-3 \tan 2 A}$

Step 7:

= 3.tan3A   (Equal to RHS)

Therefore  LHS=RHS

Hence it is satisfied

Answer 2

Step-by-step explanation:

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