Show that the equation tan(30◦ + θ) = 2 tan(60◦ − θ) can be written in the form tan2 θ + (6 √3)tan θ − 5 = 0.?

Identities:
tan(A + B) = (tanA + tanB) / (1 - tanA tanB)
tan(A - B) = (tanA - tanB) / (1 + tanA tanB)

Special angles :
tan30° = 1/√3 and tan60° = √3

tan(30° + θ) = 2 tan(60° - θ)
(tan30° + tanθ) / (1 - tan30° tanθ) = 2 (tan60° - tanθ) / (1 + tan60° tanθ)
[(1/√3) + tanθ] / [1 - (1/√3) tanθ] = 2 (√3 - tanθ) / (1 + √3 tanθ)
{[(1/√3) + tanθ] × √3} / {[1 - (1/√3) tanθ] × √3} = 2 (√3 - tanθ) / (1 + √3 tanθ)
(1 + √3 tanθ) / (√3 - tanθ) = 2 (√3 - tanθ) / (1 + √3 tanθ)
(1 + √3 tanθ)² = 2 (√3 - tanθ)²
1 + 2 √3 tanθ + 3 tan²θ = 2 (3 - 2 √3 tanθ + tan²θ)
1 + 2 √3 tanθ + 3 tan²θ = 6 - 4 √3 tanθ + 2 tan²θ
(3 tan²θ - 2 tan²θ) + (2 √3 tanθ + 4 √3 tanθ) + (1 - 6) = 0
tan²θ + 6 √3 tanθ - 5 = 0

Hence, tan²θ + 6 √3 tanθ - 5 = 0

i) tan(60 - θ) = tan{90 - (60 + θ)} = cot(30 + θ)

ii) So, given expression is: tan(30 + θ) = 2*cot(30 + θ) = 2/tan(30 + θ)
==> tan²(30 + θ) = 2
[(tan30 + tanθ)/(1 - tan30*tanθ)]² = 2

==> (1 + √3 tanθ)²/(√3 - tanθ)² = 2

==> 1 + 2 √3 tanθ + 3tan²θ = 2(3 - 2 √3 tanθ + tan²θ)

This simplifies as: tan²θ + 6√3 tanθ - 5 = 0

tan(α+β) = (tan(α) + tan(β)) / (1 - tan(α)tan(β))
tan(α-β) = (tan(α) - tan(β)) / (1 + tan(α)tan(β))

tan(30°+θ) = (tan(30°) + tan(θ)) / (1 - tan(30°)tan(θ))
= (1/√3 + tan(θ)) / (1 - (1/√3)tan(θ))

tan(60°-θ) = (tan(60°) - tan(θ))/(1 + tan(60°)tan(θ))
= (√3 - tan(θ))/(1 + (√3)tan(θ))

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Now, consider tan(30°+θ) - 2tan(60°-θ) = 0.

Expand this as
(1/√3 + tan(θ)) / (1 - (1/√3)tan(θ)) - 2(√3 - tan(θ)) / (1 + (√3)tan(θ)) = 0.

Clear the denominators by multiplying both sides by (1 - (1/√3)tan(θ))·(1 + (√3)tan(θ)), giving
(1/√3 + tan(θ))·(1 + (√3)tan(θ)) - 2(√3 - tan(θ))·(1 - (1/√3)tan(θ)) = 0.

By FOIL, this is
(1/√3 + tan(θ) + tan(θ) + (√3)tan²(θ)) - 2(√3 - tan(θ) - tan(θ) + (1/√3)tan²(θ)) = 0.

Collect terms and distribute the 2 over the second portion of the left-hand side, giving
1/√3 + 2tan(θ) + (√3)tan²(θ) - 2√3 + 4tan(θ) - (2/√3)tan²(θ)) = 0.

-2√3 = -2(3/√3). Combine this with 1/√3 and collect the tan(θ) terms, giving
-5/√3 + 6tan(θ) + (1/√3)tan²(θ) = 0.

Multiply through by √3, giving
-5 + (6√3)tan(θ) + tan²(θ) = 0, or
tan²(θ) + (6√3)tan(θ) - 5 = 0.