Solution:

Given, sin θ sin²θ = 1

Let's simplify the given equation:

sin θ sin²θ = 1

sin θ (sin θ)² = 1

(sin²θ)² = 1

sin⁴θ = 1

Taking square root on both sides, we get

sin²θ = ±1

But sin²θ cannot be negative, as it is a square of a real number. Therefore, sin²θ = 1

Now, we have to find the value of cos¹²θ + 3cos10θ + 3cos8θ + cos6θ + 2cos⁴θ + 2cos²θ - 2

Let's simplify each term one by one:

cos¹²θ = (cos²θ)⁶ = (1 - sin²θ)⁶ = (1 - 1)⁶ = 0

3cos10θ = 3(cos²θ)⁵sinθ

= 3(1 - sin²θ)²sinθ

= 3sinθ

= 3√(1 - cos²θ)

= 3√(1 - (1 - sin²θ))

= 3√sin²θ

= 3sinθ

= 3

3cos8θ = 3(cos²θ)⁴(sin²θ) = 3(1 - sin²θ)²(sin²θ)

= 3sin²θ(1 - sin²θ)

= 3(1 - cos²θ)(sin²θ)

= 3sin²θ - 3sin⁴θ

= 3 - 3sin²θ

= 2

cos6θ = cos²(3θ) = (1 - sin²(3θ))

= (1 - 4sin²θ + 3sin⁴θ)

2cos⁴θ = 2(cos²θ)² = 2(1 - sin²θ)²

= 2(1 - 2sin²θ + sin⁴θ)

= 2 - 4sin²θ + 2sin⁴θ

2cos²θ = 2(1 - sin²θ)

- 2

Now, substituting the values in the given expression, we get:

cos¹²θ + 3cos10θ + 3cos8θ + cos6θ + 2cos⁴θ + 2cos²θ - 2

= 0 + 3 + 2 + (1 - 4sin²θ + 3sin⁴θ) + (2 - 4sin²θ + 2sin⁴θ) + (2 - 2sin²θ) - 2

= 9

Hence, the value of the expression is 9.

Therefore, the solution is as follows:

Solution:

Given: sin θ sin²θ = 1

To find: Value of cos¹²θ + 3cos10θ + 3cos8θ + cos6θ + 2cos⁴θ + 2cos²θ - 2

Steps to solve the problem:

1. Simplify sin θ sin²θ = 1