If $$\text{cos}^4\text{θ}-\text{sin}^4\text{θ}=\frac{2}{3}$$​ then the value of $$1-2\text{sin}^2\text{θ}$$​ is:

Correct option is D

Given:

​$$\cos^4\theta - \sin^4\theta = \frac{2}{3}$$​​

We are to find the value of:

​$$1 - 2\sin^2\theta$$​

Concept Used:
Use the identity:

​$$a^4 - b^4 = (a^2 + b^2)(a^2 - b^2)$$​

Also,

​$$\sin^2\theta + \cos^2\theta = 1$$​

Solution:

​$$\cos^4\theta - \sin^4\theta = (\cos^2\theta + \sin^2\theta)(\cos^2\theta - \sin^2\theta)$$ =1. $$(\cos^2\theta - \sin^2\theta)$$​​

Given:

​$$\cos^4\theta - \sin^4\theta = \frac{2}{3}$$​​

​$$\cos^2\theta - \sin^2\theta = \frac{2}{3}$$​​

​$$(1 - \sin^2\theta) - \sin^2\theta = \frac{2}{3}$$​

​$$1 - 2\sin^2\theta = \frac{2}{3}$$

Alternate Method :

Let x $$= \sin^2\theta$$​​

Then:

​$$\cos^2\theta = 1 - x$$​​

Now:

​$$\cos^4\theta - \sin^4\theta = (1 - x)^2 - x^2 = 1 - 2x + x^2 - x^2 = 1 - 2x$$​​

Given:

​$$1 - 2x = \frac{2}{3}$$​

​$$x = \frac{1}{6}$$​ 

$$\sin^2\theta = \frac{1}{6}$$​

​$$1 - 2\sin^2\theta = 1 - 2 \times \frac{1}{6} = \frac{2}{3}$$​