Answer: C
\( \tan^4\theta + \tan^2\theta = 1 \) i.e., \( \tan^2\theta(\tan^2\theta + 1) = 1 \) i.e., \( \tan^2\theta \times \sec^2\theta = 1 \) i.e., \( \cfrac{\sin^2\theta}{\cos^2\theta} \times \cfrac{1}{\cos^2\theta} = 1 \) i.e., \( \cos^4\theta = \sin^2\theta \) \(\therefore \cos^4\theta + \cos^2\theta - 1 \) \(= \sin^2\theta + \cos^2\theta - 1 \) [since \( \cos^4\theta = \sin^2\theta \)] \(= 1 - 1 = 0 \)
\( \tan^4\theta + \tan^2\theta = 1 \) i.e., \( \tan^2\theta(\tan^2\theta + 1) = 1 \) i.e., \( \tan^2\theta \times \sec^2\theta = 1 \) i.e., \( \cfrac{\sin^2\theta}{\cos^2\theta} \times \cfrac{1}{\cos^2\theta} = 1 \) i.e., \( \cos^4\theta = \sin^2\theta \) \(\therefore \cos^4\theta + \cos^2\theta - 1 \) \(= \sin^2\theta + \cos^2\theta - 1 \) [since \( \cos^4\theta = \sin^2\theta \)] \(= 1 - 1 = 0 \)