The value of $\frac{1-2 \sin ^2 \theta \cos ^2 \theta}{\sin ^4 \theta+\cos ^4 \theta}-1$ is:

0

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

= \(\frac{1 - 2 sin²θ . cos²θ}{(sin²θ)² + (cos²θ)² }\) - 1

= \(\frac{1 - 2 sin²θ . cos²θ}{(sin²θ + cos²θ)²  - 2 sin²θ . cos²θ }\) - 1

{ we know, sin²θ + cos²θ = 1 } - 1

= \(\frac{1 - 2 sin²θ . cos²θ}{1  - 2 sin²θ . cos²θ }\) - 1

= 0