Answer :

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.