If $\alpha$ and $\beta$ are the roots of the quadratic equation, $x^{2}+x \sin \theta-2 \sin \theta=0, \theta \in\left(0, \frac{\pi}{2}\right)$, then
$\frac{\alpha^{12}+\beta^{12}}{\left(\alpha^{-12}+\beta^{-12}\right)(\alpha-\beta)^{24}}$ is equal to :
Correct Option: , 4
$\frac{\alpha^{12}+\beta^{12}}{\left(\frac{1}{\alpha^{12}}+\frac{1}{\beta^{12}}\right)(\alpha-\beta)^{24}}=\frac{(\alpha \beta)^{12}}{(\alpha-\beta)^{24}}$
$=\frac{(\alpha \beta)^{12}}{\left[(\alpha+\beta)^{2}-4 \alpha \beta\right]^{12}}=\left[\frac{\alpha \beta}{(\alpha+\beta)^{2}-4 \alpha \beta}\right]^{12}$
$=\left(\frac{-2 \sin \theta}{\sin ^{2} \theta+8 \sin \theta}\right)^{12}=\frac{2^{12}}{(\sin \theta+8)^{12}}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.