Given: 2tanA = 3tanB
Divide both sides by 2tanB:
tanA / tanB = 3/2
Now, using the identity tan(A+B) = (tanA + tanB) / (1 - tanA*tanB), we can substitute tanA/tanB = 3/2:
tan(A+B) = (tanA + tanB) / (1 - tanA*tanB)
tan(A+B) = (3tanB + tanB) / (1 - 3/2)
tan(A+B) = 4tanB / (1/2)
tan(A+B) = 8tanB
Since tan2θ = 2tanθ / (1 - tan²θ), substitute tanB for θ:
tan2B = 2tanB / (1 - tan²B)
tan2B = 2tanB / (1 + tan²B)
tan2B(1 + tan²B) = 2tanB
tan²B + tan²Btan²B = 2tanB
tan²B(1 + tan²B) = 2tanB
tan²B(1 + tan²B) - 2tanB = 0
tan²B + tan⁴B - 2tanB = 0
Now, let's solve this quadratic equation by substituting tanB with x:
x² + x⁴ - 2x = 0
x(x³ + 1) - 2(x) = 0
x(x³ - 2) + (x³ - 2) = 0
(x + 1)(x³ - 2) = 0
x = -1 or x³ = 2
As tanB must be positive, the only possible value is x = ∛2.
Therefore, tanB = ∛2
Now substitute tanB back into the expression for tan(A+B):
tan(A+B) = 8∛2
To simplify this expression, use the double angle formula: sin2θ = 2sinθcosθ and cos2θ = 2cos²θ - 1
sin2B = 2sinBcosB and cos2B = 2cos²B - 1
sin2B = 2√(1 - cos²B)cosB and cos2B = 2cos²B - 1
Substitute into the expression for tan(A+B):
tan(A+B) = 8∛2
= 8(2√(1 - cos²B)cosB)/(2cos²B - 1)
= (16√(1 - cos²B)cosB)/(2cos²B - 1)
Therefore, tan(A+B) = (16√(1 - cos²B)cosB)/(2cos²B - 1)
= (4√(1 - cos²B)cosB)/(cos²B - 1/2)
Hence, tan(A+B) = (5sin2B)/(5cos2B-1) as required.
