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Step-by-Step Solution
Step 1: Identify the sum and product of the roots
For the quadratic equation
$x^2 + x\,\sin\theta - 2\,\sin\theta = 0,$
let its roots be $ \alpha $ and $ \beta $. By Viète’s formulas:
$ \alpha + \beta = -\sin\theta \quad\text{and}\quad \alpha\,\beta = -2\,\sin\theta. $
Step 2: Express the given ratio in simpler form
We want to find the value of
$$
\frac{\alpha^{12} + \beta^{12}}
{\bigl(\alpha^{-12} + \beta^{-12}\bigr)\,\bigl(\alpha - \beta\bigr)^{24}}.
$$
Observe that
$$
\alpha^{-12} + \beta^{-12}
= \frac{1}{\alpha^{12}} + \frac{1}{\beta^{12}}
= \frac{\alpha^{12} + \beta^{12}}{\alpha^{12}\,\beta^{12}}.
$$
Thus, the denominator
$
\bigl(\alpha^{-12} + \beta^{-12}\bigr)\,\bigl(\alpha - \beta\bigr)^{24}
$
can be written as
$$
\frac{\alpha^{12} + \beta^{12}}{\alpha^{12}\,\beta^{12}}\times (\alpha - \beta)^{24}.
$$
Hence the given ratio simplifies to
$$
\frac{\alpha^{12} + \beta^{12}}{\bigl(\alpha^{-12} + \beta^{-12}\bigr)\,(\alpha - \beta)^{24}}
= \frac{\alpha^{12} + \beta^{12}}
{\displaystyle \frac{\alpha^{12} + \beta^{12}}{\alpha^{12}\,\beta^{12}}\,(\alpha - \beta)^{24}}
= \frac{\alpha^{12}\,\beta^{12}}{(\alpha - \beta)^{24}}.
$$
Therefore, the expression reduces to
$
\dfrac{(\alpha\,\beta)^{12}}{(\alpha - \beta)^{24}}.
$
Step 3: Calculate $(\alpha - \beta)^2$
Using the identity
$
(\alpha - \beta)^2
= (\alpha + \beta)^2 - 4\,\alpha\,\beta,
$
we substitute $ \alpha + \beta = -\sin\theta $ and $ \alpha\,\beta = -2\,\sin\theta $:
$$
(\alpha - \beta)^2
= (-\sin\theta)^2 - 4(-2\,\sin\theta)
= \sin^2\theta + 8\,\sin\theta.
$$
Hence,
$
\alpha - \beta
= \sqrt{\sin^2\theta + 8\,\sin\theta}
= \sqrt{\sin\theta\,(\sin\theta + 8)}.
$
Therefore,
$
(\alpha - \beta)^{24}
= \bigl[(\alpha - \beta)^2\bigr]^{12}
= \bigl[\sin\theta\,(\sin\theta + 8)\bigr]^{12}
= \sin^{12}\theta \,(\sin\theta + 8)^{12}.
$
Step 4: Substitute into the reduced expression
Since
$
\alpha\,\beta = -2\,\sin\theta,
$
its magnitude for our expression is $2\,\sin\theta$. Thus,
$$
\frac{(\alpha\,\beta)^{12}}{(\alpha - \beta)^{24}}
= \frac{\bigl(2\,\sin\theta\bigr)^{12}}
{\sin^{12}\theta\,(\sin\theta + 8)^{12}}
= \frac{2^{12}\,\sin^{12}\theta}
{\sin^{12}\theta\,(\sin\theta + 8)^{12}}
= \frac{2^{12}}{(\sin\theta + 8)^{12}}.
$$
Step 5: State the final answer
Therefore, the value of
$
\frac{\alpha^{12} + \beta^{12}}
{\bigl(\alpha^{-12} + \beta^{-12}\bigr)\,\bigl(\alpha - \beta\bigr)^{24}}
$
is
$$
\boxed{\frac{2^{12}}{(\sin\theta + 8)^{12}}}.
$$
