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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