Question

I have doubt how
tan3Atan2AtanA=tan3A-tan2A-tanA
I try. I did not get!

Answer 1

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please refer to the attachment above


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Answer 2

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Given,

Tan3Atan2AtanA=tan3A-tan2A-tanA

That,

3A= 2A+A

$tan3a = \frac{ \tan(2a) + \tan(a) }{1 - \tan(2a) \tan(a) }$

tan-3A-tan3Atan2AtanA=tan2A+tanA

So

LHS=RHS

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