Prove that:tan x+tan(60+x)+tan(120+x)=3
Answers
Answer:
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
Answer:
Answer:
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