Question

Prove that: tan 3A – tan 2A – tan A = tan 3A tan 2A tan A​

Answer 1

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Question :  Prove that: tan 3A – tan 2A – tan A = tan 3A tan 2A tan A​

Answer : tan 3A – tan 2A – tan A = tan 3A tan 2A tan A​

Formula :   $tan(A+B)=\frac{tanA+tanB}{1-tanA.tanB}$

Step-by-step explanation :

$tan(3A)=tan(2A+A)\\\\=>tan(3A)=\frac{tan2A+tanA}{1-tan2A.tanA}\\\\ =>tan(3A)(1-tan2A.tanA)=tan2A+tanA\\\\=>tan3A-tan3A.tan2A.tanA=tan2A+tanA\\\\=>tan3A-tan2A-tanA=tan3A.tan2A.tanA$

Hence Proved .

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Answer 2

Given :   tan 3A – tan 2A – tan A = tan 3A tan 2A tan A​

To find : Prove  

Solution:

Tan3A = Tan(2A  + A)

as we know that Tan(X + Y)   = (TanX  + TanY) /(1 - TanXTanY)

X = 2A  & Y = A

=> Tan3A =   (Tan2A + TanA)/(1 - Tan2A.TanA)

=> (1 - Tan2A.TanA)Tan3A = Tan2A + TanA

=> Tan3A - Tan3A.Tan2A.TanA  = Tan2A + TanA

=> Tan3A - Tan2A - TanA = Tan3A.Tan2A.TanA

QED

Hence Proved

Tan3A - Tan2A - TanA = Tan3A.Tan2A.TanA

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