Question

Prova that
tanA × Tan (60°+A) × Tan (120°+A) = -tan3A
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Answer 1

Step-by-step explanation:

tanA × Tan (60°+A) × Tan (120°+A)

Tan (60°+A) =(tan60+tanA)/(1-tan60tanA)

=(root3+tanA)/(1-root3tanA)

Tan (120°+A) =(tan120+tanA)/(1-tan120tanA)

=(-root3+tamA)/(1+root3tanA)

since tan120=tan(180-60)=-tan60= -root3

tanA × Tan (60°+A) × Tan (120°+A)

=tanA × (root3+tanA)/(1-root3tanA) × (-root3+tamA)/(1+root3tanA)

=tanA(tan^2A - (root3)^2)/(1^2-(root3tanA)^2)

= (tan^3A-3tanA)/(1-3tan^2A)

= -(3tanA-tan^3A)/(1-3tan^2A)

= -tan3A

Answer 2

See above answer.......

Thanks...... for points ✌