Find the value of $$1 - 2 sin^{2} θ + sin^{4} θ.$$

Here,

$$1 - 2 sin^{2} θ + sin^{4} θ.$$ = $$1^2 + (sin^2 \theta)^2 - 2 \times 1 \times sin^2 \theta$$

it is similar to $$(a-b)^2$$ = $$a^2 + b^2 - 2ab$$

So,

$$1^2 + (sin^2 \theta)^2 - 2 \times 1 \times sin^2 \theta$$ = $$(sin^2 \theta - 1)^2$$......(1)

Now $$sin^2 \theta + cos^2 \theta$$ = 1.....(2)

From equation 1 and 2

$$(sin^2 \theta - 1)^2$$ = $$(sin^2 \theta - sin^2 \theta - cos^2 \theta)^2$$

= $$(cos^2 \theta)^2$$

= $$cos^4 \theta$$