If $\alpha $and$\beta $are the roots of the quadratic equation, ${x^2} + x\sin \theta - 2\sin \theta = 0,\theta \in (0,\dfrac{\pi }{2}),$then
$\dfrac{{{\alpha ^{12}} + {\beta ^{12}}}}{{({\alpha ^{ - 12}} + {\beta ^{ - 12}}){{(\alpha - \beta )}^{24}}}}$is equal to:
$
  {\text{A}}{\text{. }}\dfrac{{{2^6}}}{{{{(\sin \theta + 8)}^{12}}}}{\text{ }} \\
  {\text{B}}{\text{.}}\dfrac{{{2^{12}}}}{{{{(\sin \theta - 8)}^6}}}{\text{ }} \\
  {\text{C}}{\text{.}}\dfrac{{{2^{12}}}}{{{{(\sin \theta - {\text{4}})}^{12}}}} \\
  {\text{D}}{\text{. }}\dfrac{{{2^{12}}}}{{{{(\sin \theta {\text{ + 8}})}^{12}}}} \\
 $function as

Hint:-In this problem first we find the relation between roots and coefficients of the given quadratic equation{${x^2} + x\sin \theta - 2\sin \theta = 0,\theta \in (0,\dfrac{\pi }{2})$}. Then, we simplify the given function till it comes in the form of relation between roots and coefficients of the given quadratic and then put their values in it to get the answer.

Complete step-by-step answer:
We know the relation between roots and the coefficients of the quadratic equation in the form of $a{x^2} + bx + c = 0$. Let roots of$a{x^2} + bx + c = 0$be $\alpha $and$\beta $then relation is given by:
Sum of roots$(\alpha + \beta )$=$\dfrac{{ - b}}{a}$ eq.1
And product of roots=$\dfrac{c}{a}$ eq.2
Now, it is given that If $\alpha $and$\beta $are the roots of the quadratic equation, ${x^2} + x\sin \theta - 2\sin \theta = 0,\theta \in (0,\dfrac{\pi }{2}),$then using eq.1 and eq.2 we get
$
   \Rightarrow \alpha + \beta = - \sin \theta {\text{ eq}}{\text{.3}} \\
  {\text{and,}} \\
   \Rightarrow \alpha \times \beta = - 2\sin \theta {\text{ eq}}{\text{.4 }} \\
 $
Now, consider the$\dfrac{{{\alpha ^{12}} + {\beta ^{12}}}}{{({\alpha ^{ - 12}} + {\beta ^{ - 12}}){{(\alpha - \beta )}^{24}}}}$
We know ${a^{ - m}} = \dfrac{1}{{{a^m}^{}}}$
Then, above equation can be written as
$ \Rightarrow \dfrac{{{\alpha ^{12}} + {\beta ^{12}}}}{{(\dfrac{1}{{{\alpha ^{12}}}} + \dfrac{1}{{{\beta ^{12}}}}){{(\alpha - \beta )}^{24}}}}$
On taking LCM in denominator we can rewrite above equation as
$
   \Rightarrow \dfrac{{{\alpha ^{12}} + {\beta ^{12}}}}{{\{ \dfrac{{({\alpha ^{12}} + {\beta ^{12}})}}{{{{(\alpha \beta )}^{12}}}}\} {{(\alpha - \beta )}^{24}}}} \\
   \Rightarrow \dfrac{{{{(\alpha \beta )}^{12}}}}{{{{(\alpha - \beta )}^{24}}}}{\text{ eq}}{\text{.5}} \\
 $
We know, ${(a - b)^2} = {(a + b)^2} - 4ab$
Using above property we can rewrite eq.5 as
$
   \Rightarrow \dfrac{{{{(\alpha \beta )}^{12}}}}{{{{\{ {{(\alpha + \beta )}^2} - 4\alpha \beta \} }^{12}}}} \\
   \Rightarrow {\left[ {\dfrac{{\alpha \beta }}{{{{(\alpha + \beta )}^2} - 4\alpha \beta }}} \right]^{12}}{\text{ eq}}{\text{.6}} \\
  {\text{ }} \\
 $
Now put values of $\alpha + \beta $and$\alpha \times \beta $into eq.6 , we get
$
   \Rightarrow {\left[ {\dfrac{{ - 2\sin \theta }}{{{{\sin }^2}\theta - 4( - 2\sin \theta )}}} \right]^{12}} \\
   \Rightarrow {\left[ {\dfrac{{ - 2\sin \theta }}{{{{\sin }^2}\theta {\text{ + 8(}}\sin \theta )}}} \right]^{12}} \\
$
In the above equation $\sin \theta $is common in both numerator and denominator so cancel it.
We get
$
   \Rightarrow {\left[ {\dfrac{2}{{\sin \theta {\text{ + 8}}}}} \right]^{12}}{\text{ \{ ( - 1}}{{\text{)}}^{2n}} = 1\} \\
   \Rightarrow \dfrac{{{2^{12}}}}{{{{(\sin \theta + 8)}^{12}}}} \\
$
Hence, option D is correct.

Note:Whenever you get this type of question the key concept of solving this is to just simply the given equation as much as possible by using properties of trigonometric angles and relation between the roots of quadratic equation($a{x^2} + bx + c = 0$) and its coefficients to get the result in simplest form.