If cosθ+cos² θ=1, prove that

sin¹² θ+3sin¹⁰ θ+3sin⁸ θ+sin⁶ θ+2sin⁴ θ+2sin² θ-2=1

Proof:

As cosθ+cos² θ=1
cos θ =1-cos² θ
cos θ =sin² θ ….(i)

Using(a + b)³ =a³+b³+3(a + b)
Now,sin¹² θ+3sin¹⁰ θ+3sin⁸ θ+sin⁶ θ+2sin⁴ θ+2sin² θ-2= (sin⁴ θ)³+sin⁴ θ-sin² θ[sin⁴ θ+sin² θ](sin² θ)³+2(sin² θ)²+2sin² θ-2
=(sin⁴ θ+sin² θ)³+2(cos θ)²+2cosθ-2.
=((sin² θ)²+sin² θ)³ +2cos² θ+2cosθ-2

From (1),
=(cos² θ +sin² θ)³+2cos² θ+2cosθ-2

As sin²θ + cos²θ = 1
= 1 +2cos² θ+2sin² θ-2
= 1 +2(cos² θ+sin² θ)-2
=1+2(1)-2

=1

Hence proved.