The second member of the sum is a product between the square tangent of an angle and the square cosine of the same angle.

Remember that `tan alpha = (sin alpha)/(cos alpha)` => `tan^2 alpha = (sin^2 alpha)/(cos^2 alpha)`

Use this relation to simplify the product `tan^2(3pi/5)*cos^2(3pi/5) = (sin^2(3pi/5))*(cos^2(3pi/5))/(cos^2(3pi/5)) = sin^2(3pi/5)`

Write the sum replacing the second product by `sin^2(3pi/5).` `sin(2pi/5)*cos(3pi/5) + sin^2(3pi/5).`

Write `sin (2pi/5) = sin (pi - 3pi/5) = cos (3pi/5)`

Substitute `sin(2pi/5)` by `cos (3pi/5).`

Write the sum including the substituted factor.

`cos (3pi/5)*cos (3pi/5) + sin^2(3pi/5)`

`` `cos^2 (3pi/5) + sin^2(3pi/5)`

Remember the basic formula of trigonometry:

`sin^2 alpha + cos^2 alpha = 1`

Put `alpha = 3pi/5`

`cos^2 (3pi/5) + sin^2(3pi/5) = 1`

**ANSWER: Using the complementary angles properties and basic identities of trigonometry, the sum `sin(2pi/5)*cos(3pi/5) + tan^2(3pi/5)*cos^2(3pi/5) = 1.` **

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