The value of the limit $$\lim_{\theta \to 0} \frac{\tan(\pi\cos^2\theta)}{\sin(2\pi\sin^2\theta)}$$ is equal to:

We need to evaluate $$\lim_{\theta \to 0} \frac{\tan(\pi\cos^2\theta)}{\sin(2\pi\sin^2\theta)}$$.

As $$\theta \to 0$$, we have $$\cos^2\theta \to 1$$ and $$\sin^2\theta \to 0$$. So the argument of $$\tan$$ approaches $$\pi$$ and the argument of $$\sin$$ approaches $$0$$.

We write $$\cos^2\theta = 1 - \sin^2\theta$$, so $$\pi\cos^2\theta = \pi - \pi\sin^2\theta$$.

Using the identity $$\tan(\pi - x) = -\tan(x)$$, we get $$\tan(\pi\cos^2\theta) = \tan(\pi - \pi\sin^2\theta) = -\tan(\pi\sin^2\theta)$$.

The limit becomes $$\lim_{\theta \to 0} \frac{-\tan(\pi\sin^2\theta)}{\sin(2\pi\sin^2\theta)}$$.

Let $$t = \sin^2\theta$$. As $$\theta \to 0$$, $$t \to 0$$. The expression becomes $$\lim_{t \to 0} \frac{-\tan(\pi t)}{\sin(2\pi t)}$$.

For small $$t$$, $$\tan(\pi t) \approx \pi t$$ and $$\sin(2\pi t) \approx 2\pi t$$. Therefore:

$$\lim_{t \to 0} \frac{-\tan(\pi t)}{\sin(2\pi t)} = \lim_{t \to 0} \frac{-\pi t}{2\pi t} = -\frac{1}{2}$$.

The answer is $$-\frac{1}{2}$$, which is Option A.