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If a=tanA b=tan2A and c=tan3A what is the value of (c-b-a)/abc?

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Given,

a=tanA, b=tan2A and c=tan3A

3A=2A+A

⇒tan3A=tan(2A+A)=(tan2A+tanA)/(1−tan2AtanA)

⇒tan3A−tan3Atan2AtanA=tan2A+tanA

⇒tan3A−tan2A−tanA=tan3Atan2AtanA

Hence, (c-b-a)/abc = 1

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Deependra Verma

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