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3 years ago

Please someone help me to prove this.

Mathematics

2 answers:

kodGreya [7K]3 years ago

5 0

Hope this can help.

KengaRu [80]3 years ago

4 0

Answer: see proof below

Step-by-step explanation:

Use the following Sum Identity:

$\tan (A + B) =\dfrac{\tan A+\tan B}{1-\tan A\cdot \tan B}$

Given: A + B + C = 180° → A + B + C = π

Proof LHS → RHS

Given: A + B + C = π

Multiply by 2: 2(A + B + C = π)

→ 2A + 2B + 2C = 2π

→ 2A + 2B = 2π - 2C

Apply tan: tan(2A + 2B = 2π - 2C)

→ tan (2A + 2B) = tan(2π - 2C)

→ tan (2A + 2B) = - tan 2C

$\text{Sum Identity:}\qquad \qquad \dfrac{\tan 2A+\tan 2B}{1-\tan 2A\cdot \tan 2B}=-\tan 2C$

Simplify: tan 2A + tan 2B = -tan 2C (1 - tan 2A · tan 2B)

Distribute: tan 2A + tan 2B = -tan 2C + tan 2A · tan 2B · tan 2C

Add tan 2C: tan 2A + tan 2B + tan 2C = tan 2A · tan 2B · tan 2C

LHS = RHS is proven

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Step 1:

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