Math, asked by VirajSingh11, 1 year ago

please solve this question​

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Answered by BrainlyTornado
3

CORRECT QUESTION:

Prove that Tan 8θ - Tan 6θ - Tan 2θ = Tan 8θ Tan 6θ Tan 2θ.

GIVEN:

Tan 8θ - Tan 6θ - Tan 2θ = Tan 8θ Tan 6θ Tan 2θ.

TO PROVE:

Tan 8θ - Tan 6θ - Tan 2θ = Tan 8θ Tan 6θ Tan 2θ.

PROOF:

Let L.H.S = Tan 8θ - Tan 6θ - Tan 2θ.

Let R.H.S = Tan 8θ Tan 6θ Tan 2θ.

L.H.S:

Tan 8θ - (Tan 6θ + Tan 2θ)

\boxed{\bold{\large{Tan(A+B) = \dfrac{Tan\ A + Tan\ B}{1-Tan\ A \ Tan\ B}}}}

(Tan A + B)(1 - Tan A Tan B) = Tan A + Tan B

Take A = 6θ and B = 2θ [ As they are in the form of Tan A + Tan B ]

(Tan 6θ + 2θ)(1 - Tan 6θ Tan 2θ) = Tan 6θ + Tan 2θ

(Tan 8θ)(1 - Tan 6θ Tan 2θ) = Tan 6θ + Tan 2θ

Substitute the value of Tan 6θ + Tan 2θ

Tan 8θ - (Tan 8θ)(1 - Tan 6θ Tan 2θ)

Take Tan 8θ as common

Tan 8θ(1 - (1 - Tan 6θ Tan 2θ))

Tan 8θ(1 - 1 + Tan 6θ Tan 2θ)

Tan 8θ Tan 6θ Tan 2θ

R.H.S:

Tan 8θ Tan 6θ Tan 2θ.

L.H.S = R.H.S

Tan 8θ - Tan 6θ - Tan 2θ = Tan 8θ Tan 6θ Tan 2θ.

HENCE PROVED.

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