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JEE MAIN_2019_
10-04-2019-S-1
Question
If $\alpha$ and $\beta$ are the roots of the quadratic equation, $x^2+x \sin \theta-2 \sin \theta=0, \quad \theta \in\left(0, \frac{\pi}{2}\right)$, then $\frac{\alpha^{12}+\beta^{12}}{\left(a^{-12}+\beta^{-12}\right) \cdot(\alpha-\beta)^{24}}$ is equal to :
Select the correct option:
A
$\frac{2^{12}}{(\sin \theta-8)^5}$
B
$\frac{2^{12}}{(\sin \theta-4)^{12}}$
C
$\frac{2^6}{(\sin \theta+8)^{12}}$
D
$\frac{2^{12}}{(\sin \theta+8)^{12}}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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Question Tags
JEE Main
Mathematics
Medium
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