If $$\sin^4 \theta - \cos^4 \theta = \frac{1}{2}$$, find the value of $$2 \sin^2 \theta - 1$$.​

Correct option is B

Given: ​

​$$\sin^4 \theta - \cos^4 \theta = \frac{1}{2}$$ 

Formula Used: ​

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

Solution: ​

​$$\sin^4 \theta - \cos^4 \theta = \frac{1}{2} \\ \ \\ (\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta) = \frac{1}{2} \\ \ \\ \sin^2 \theta - \cos^2 \theta = \frac{1}{2} \\ \ \\ \sin^2 \theta - (1 -\sin^2 \theta) = \frac{1}{2} \\ \ \\ 2\sin^2 \theta -1 = \frac{1}{2}$$ ​