1. The problem is to simplify the expression $$1 + \frac{\tan^2(\alpha) - 1}{\sin\left(0.5\pi + 2\alpha\right)}.$$ 2. Recall the Pythagorean identity: $$\tan^2(\alpha) + 1 = \sec^2(\alpha),$$ so $$\tan^2(\alpha) - 1 = \sec^2(\alpha) - 2.$$ However, it is easier to use the identity $$\tan^2(\alpha) = \sec^2(\alpha) - 1,$$ so $$\tan^2(\alpha) - 1 = \sec^2(\alpha) - 2.$$ 3. Also, use the sine addition formula and periodicity: $$\sin\left(0.5\pi + 2\alpha\right) = \cos(2\alpha).$$ 4. Substitute these into the expression: $$1 + \frac{\tan^2(\alpha) - 1}{\sin\left(0.5\pi + 2\alpha\right)} = 1 + \frac{\tan^2(\alpha) - 1}{\cos(2\alpha)}.$$ 5. Using the double angle identity for cosine: $$\cos(2\alpha) = 1 - 2\sin^2(\alpha) = 2\cos^2(\alpha) - 1.$$ 6. Rewrite numerator using $$\tan^2(\alpha) = \frac{\sin^2(\alpha)}{\cos^2(\alpha)}$$: $$\tan^2(\alpha) - 1 = \frac{\sin^2(\alpha)}{\cos^2(\alpha)} - 1 = \frac{\sin^2(\alpha) - \cos^2(\alpha)}{\cos^2(\alpha)}.$$ 7. Substitute back: $$1 + \frac{\frac{\sin^2(\alpha) - \cos^2(\alpha)}{\cos^2(\alpha)}}{\cos(2\alpha)} = 1 + \frac{\sin^2(\alpha) - \cos^2(\alpha)}{\cos^2(\alpha) \cos(2\alpha)}.$$ 8. Note that $$\sin^2(\alpha) - \cos^2(\alpha) = -\cos(2\alpha),$$ so $$1 + \frac{-\cos(2\alpha)}{\cos^2(\alpha) \cos(2\alpha)} = 1 - \frac{1}{\cos^2(\alpha)}.$$ 9. Simplify: $$1 - \sec^2(\alpha) = 1 - (1 + \tan^2(\alpha)) = -\tan^2(\alpha).$$ 10. Therefore, the simplified expression is $$-\tan^2(\alpha).$$ 11. The correct answer is D) $$-\tg^2 \alpha.$$