If $$\alpha$$ and $$\beta$$ are the roots of the quadratic equation $$x^2 + x\sin\theta - 2\sin\theta = 0$$, $$\theta \in \left(0, \frac{\pi}{2}\right)$$, then $$\frac{\alpha^{12} + \beta^{12}}{(\alpha^{-12} + \beta^{-12})\cdot(\alpha - \beta)^{24}}$$ is equal to:

We start with the quadratic equation

$$x^{2}+x\sin\theta-2\sin\theta=0,\qquad\theta\in\left(0,\frac{\pi}{2}\right).$$

For a quadratic $$ax^{2}+bx+c=0,$$ the sum and product of its roots are given by the well-known relations

$$\alpha+\beta=-\frac{b}{a},\qquad\alpha\beta=\frac{c}{a}.$$

Here $$a=1,\;b=\sin\theta,\;c=-2\sin\theta.$$ Substituting these values we get

$$\alpha+\beta=-\sin\theta,\qquad\alpha\beta=-2\sin\theta.$$

The required expression is

$$\frac{\alpha^{12}+\beta^{12}}{\left(\alpha^{-12}+\beta^{-12}\right)\,(\alpha-\beta)^{24}}.$$

First we simplify the factor $$\alpha^{-12}+\beta^{-12}.$$ Using the property $$x^{-n}=\dfrac{1}{x^{n}},$$ we have

$$\alpha^{-12}+\beta^{-12} =\frac{1}{\alpha^{12}}+\frac{1}{\beta^{12}} =\frac{\beta^{12}+\alpha^{12}}{\alpha^{12}\beta^{12}} =\frac{\alpha^{12}+\beta^{12}}{(\alpha\beta)^{12}}.$$

Substituting this into the whole fraction gives

$$\frac{\alpha^{12}+\beta^{12}} {\bigl[\dfrac{\alpha^{12}+\beta^{12}}{(\alpha\beta)^{12}}\bigr]\, (\alpha-\beta)^{24}} \;=\; \frac{\alpha^{12}+\beta^{12}}{\alpha^{12}+\beta^{12}} \cdot \frac{(\alpha\beta)^{12}}{(\alpha-\beta)^{24}} \;=\; \frac{(\alpha\beta)^{12}}{(\alpha-\beta)^{24}}.$$

Thus all dependence on $$\alpha^{12}+\beta^{12}$$ itself has vanished, and the task is now to evaluate $$\alpha\beta$$ and $$\alpha-\beta.$$

We already know $$\alpha\beta=-2\sin\theta,$$ so

$$(\alpha\beta)^{12}=(-2\sin\theta)^{12}=2^{12}\,(\sin\theta)^{12},$$

because an even power removes the sign.

Next we compute $$\alpha-\beta.$$ Using the identity

$$(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4\alpha\beta,$$

we substitute the values we have found:

$$(\alpha-\beta)^{2} =(-\sin\theta)^{2}-4(-2\sin\theta) =\sin^{2}\theta+8\sin\theta =\sin\theta\bigl(\sin\theta+8\bigr).$$

Since we finally need $$(\alpha-\beta)^{24},$$ we raise both sides to the 12th power (note that any sign ambiguity disappears because the exponent is even):

$$(\alpha-\beta)^{24} =\bigl[(\alpha-\beta)^{2}\bigr]^{12} =\Bigl[\sin\theta\bigl(\sin\theta+8\bigr)\Bigr]^{12} =(\sin\theta)^{12}\,(\sin\theta+8)^{12}.$$

Putting everything together we get

$$\frac{(\alpha\beta)^{12}}{(\alpha-\beta)^{24}} =\frac{2^{12}(\sin\theta)^{12}} {(\sin\theta)^{12}(\sin\theta+8)^{12}} =\frac{2^{12}}{(\sin\theta+8)^{12}}.$$

This matches Option C.

Hence, the correct answer is Option 3.