Math, asked by rajtony730, 3 months ago

tan3α-tan2α-tanα=tan3αtan2αtanα

Answers

Answered by yokeshps2005

1

Answer:

tan3α=（tan2α+tanα）/（1-tan2αtanα）

∴tan2α+tanα=tan3α（1-tan2αtanα）

∴tan2α+tanα-tan3α=-tan3αtan2αtanα

∴tan3α-tan2α-tanα=tan3αtan2αtanα

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