Question

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)\left(\alpha -\beta\right)^{24}} $ is equal to :

• A. $\frac{2^6}{(\sin \theta + 8)^{12}}$
• B. $\frac{2^{12}}{(\sin \theta - 8)^{6}}$
• C. $\frac{2^{12}}{(\sin \theta - 4)^{12}}$
• D. $\frac{2^{12}}{(\sin \theta + 8)^{12}}$

Answer 1

Answer: (D)

$\frac{\alpha^{12} +\beta^{12}}{\left(\frac{1}{\alpha^{12}} + \frac{1}{\beta^{12}}\right) \left(\alpha -\beta\right)^{24}} = \frac{\left(\alpha\beta\right)^{12}}{\left(\alpha-\beta\right)^{24}} $
$ = \frac{\left(\alpha\beta\right)^{12}}{\left[\left(\alpha+\beta\right)^{2} -4 \alpha\beta\right]^{12}} = \left[\frac{\alpha\beta}{\left(\alpha+\beta\right)^{2} -4\alpha\beta}\right]^{12} $
$ = \left(\frac{-2 \sin\theta}{\sin^{2}\theta+8\sin\theta}\right)^{12} = \frac{2^{12}}{\left(\sin\theta+8\right)^{12}} $