The value of $\frac{1-2 \sin ^2 \theta \cos ^2 \theta}{\sin ^4 \theta+\cos ^4 \theta}-1$ is:
Answer & explanation
Correct answer: option 4
$\frac{1-2 \sin ^2 \theta \cos ^2 \theta}{\sin ^4 \theta+\cos ^4 \theta}-1$
= \(\frac{1 - 2 sin²θ . cos²θ}{(sin²θ)² + (cos²θ)² }\) - 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