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

Question

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

TuteeHUB · Accepted Answer

$\frac{{{2^{12}}}}{{{{({\rm{sin\theta }} - 8)}^6}}}$

TuteeHUB · Answer

$\frac{{{2^{12}}}}{{{{({\rm{sin\theta }} - 4)}^{12}}}}$

TuteeHUB · Answer

$\frac{{{2^{12}}}}{{{{({\rm{sin\theta }} + 8)}^{12}}}}$

TuteeHUB · Answer

$\frac{{{2^6}}}{{{{({\rm{sin\theta }} + 8)}^{12}}}}$

1.	If α and β are the roots of the quadratic equation, \({{\rm{x}}^2} + {\rm{x\;sin\;\theta }} - 2{\rm{\;sin\;\theta }} = 0,{\rm{\theta }} \in \left( {0,\frac{{\rm{\pi }}}{2}} \right)\) then \(\frac{{{{\rm{\alpha }}^{12}} + {{\rm{\beta }}^{12}}}}{{\left( {{{\rm{\alpha }}^{ - 12}} + {{\rm{\beta }}^{ - 12}}} \right)\cdot{{({\rm{\alpha }} - {\rm{\beta }})}^{24}}}}\) is equal to:
A.	\(\frac{{{2^{12}}}}{{{{({\rm{sin\theta }} - 4)}^{12}}}}\)
B.	\(\frac{{{2^{12}}}}{{{{({\rm{sin\theta }} + 8)}^{12}}}}\)
C.	\(\frac{{{2^{12}}}}{{{{({\rm{sin\theta }} - 8)}^6}}}\)
D.	\(\frac{{{2^6}}}{{{{({\rm{sin\theta }} + 8)}^{12}}}}\)
Answer» C. \(\frac{{{2^{12}}}}{{{{({\rm{sin\theta }} - 8)}^6}}}\)

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