Question

steps for this proof plz​

Answer 1

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Answer 2

Question:------

• Prove tan3A-tan2A - tanA = tanAtan2Atan3A

Formula used :----

• Tan(A+B) = TanA + TanB/(1-tanAtanB)
• 3a = 2a + a

Solution:-----

Let 3A = 2A + A

multiplying by tan both sides we get,

tan3A = tan(2A+A)

using above formula in RHS now we get,

$tan3A = \frac{tan2A +tanA} {1 -tanAtan2A} \\ \\ tanA + tan2A = tan3A(1 -tanAtan2A) \\ \\ tanA + tan2A =tan3A - tanAtan2Atan3A \\ \\ \\ tan3A - tanA - tan2A = tanAtan2Atan3A \:$

Hence Proved ....

(Hope it helps you)