Math, asked by lucy2019, 1 year ago

TanA.Tan(60+A).Tan(60-A)=Tan3A​

Answers

Answered by sandy1816
2

Step-by-step explanation:

tanA[tan60+tanA/1-tan60tanA] [tan60-tanA/1+tan60tanA]

=tanA[tan²60-tan²A/1-tan²60tan²A]

=tanA[3-tan²A/1-3tan²A]

=3tanA-tan³A/1-3tan²A

=tan3A [tan3A=3tanA-tan³A/1-3tan²A]

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