If `alpha and beta` the rots of the quadratic equation, `x^(2)+x sintheta-2sintheta=0,theta in(0,(pi)/(2)),` then `(alpha^(12)+beta^(12))/((alpha^(-12)+beta^(-12))(alpha-beta)^(24))` is equal to

Explanation

To solve the problem, we need to find the value of

\[ \frac{\alpha^{12} + \beta^{12}}{(\alpha^{-12} + \beta^{-12})(\alpha - \beta)^{24}} \]

where \(\alpha\) and \(\beta\) are the roots of the quadratic equation

\[ x^2 + x \sin \theta - 2 \sin \theta = 0 \]

### Step 1: Find the roots of the quadratic equation

Using the quadratic formula, the roots \(\alpha\) and \(\beta\) can be found as follows:

\[ \alpha, \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \(a = 1\), \(b = \sin \theta\), and \(c = -2 \sin \theta\).

Calculating the discriminant:

\[ b^2 - 4ac = (\sin \theta)^2 - 4(1)(-2 \sin \theta) = \sin^2 \theta + 8 \sin \theta \]

Thus, the roots are:

\[ \alpha, \beta = \frac{-\sin \theta \pm \sqrt{\sin^2 \theta + 8 \sin \theta}}{2} \]

### Step 2: Calculate \(\alpha + \beta\) and \(\alpha \beta\)

From Vieta's formulas, we have:

\[ \alpha + \beta = -\frac{b}{a} = -\sin \theta \]
\[ \alpha \beta = \frac{c}{a} = -2 \sin \theta \]

### Step 3: Find \(\alpha - \beta\)

Using the roots:

\[ \alpha - \beta = \sqrt{(\alpha + \beta)^2 - 4\alpha \beta} \]

Substituting the values:

\[ \alpha - \beta = \sqrt{(-\sin \theta)^2 - 4(-2 \sin \theta)} = \sqrt{\sin^2 \theta + 8 \sin \theta} \]

### Step 4: Calculate \(\alpha^{12} + \beta^{12}\)

Using the identity for powers of roots:

\[ \alpha^n + \beta^n = (\alpha + \beta)(\alpha^{n-1} + \beta^{n-1}) - \alpha \beta (\alpha^{n-2} + \beta^{n-2}) \]

We can compute \(\alpha^{12} + \beta^{12}\) recursively, but it is more efficient to use the relation:

\[ \alpha^{12} + \beta^{12} = (\alpha + \beta)(\alpha^{11} + \beta^{11}) - \alpha \beta (\alpha^{10} + \beta^{10}) \]

Continuing this process, we can compute \(\alpha^{12} + \beta^{12}\) in terms of \(\sin \theta\).

### Step 5: Calculate \(\alpha^{-12} + \beta^{-12}\)

Using the relation:

\[ \alpha^{-12} + \beta^{-12} = \frac{\alpha^{12} + \beta^{12}}{\alpha^{12} \beta^{12}} = \frac{\alpha^{12} + \beta^{12}}{(\alpha \beta)^{12}} \]

### Step 6: Substitute into the original expression

Now substituting everything back into the original expression:

\[ \frac{\alpha^{12} + \beta^{12}}{(\alpha^{-12} + \beta^{-12})(\alpha - \beta)^{24}} = \frac{\alpha^{12} + \beta^{12}}{\frac{\alpha^{12} + \beta^{12}}{(\alpha \beta)^{12}} \cdot (\alpha - \beta)^{24}} \]

This simplifies to:

\[ \frac{(\alpha \beta)^{12}}{(\alpha - \beta)^{24}} \]

### Step 7: Final calculation

Substituting \(\alpha \beta = -2 \sin \theta\) and \(\alpha - \beta = \sqrt{\sin^2 \theta + 8 \sin \theta}\):

\[ = \frac{(-2 \sin \theta)^{12}}{(\sqrt{\sin^2 \theta + 8 \sin \theta})^{24}} = \frac{2^{12} \sin^{12} \theta}{(\sin^2 \theta + 8 \sin \theta)^{12}} \]

### Final Result

Thus, the final result is:

\[ \frac{2^{12}}{(\sin^2 \theta + 8 \sin \theta)^{12}} \]