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यदि द्विघाती समीकरण, $x^{2}+x \sin \theta-2 \sin \theta=0, \theta \in\left(0, \frac{\pi}{2}\right) \text {, }$ के मूल $\alpha$ तथा $\beta$ हैं, तो $\frac{\alpha^{12}+\beta^{12}}{\left(\alpha^{-12}+\beta^{-12}\right)(\alpha-\beta)^{24}}$ बराबर हैं
$\frac{{{2^{12}}}}{{{{\left( {\sin \,\theta + 8} \right)}^{12}}}}$
$\frac{{{2^{12}}}}{{{{\left( {\sin \,\theta - 4} \right)}^{12}}}}$
$\frac{{{2^{12}}}}{{{{\left( {\sin \,\theta - 8} \right)}^{6}}}}$
$\frac{{{2^{6}}}}{{{{\left( {\sin \,\theta + 8} \right)}^{12}}}}$
Solution
$x^{2}+x \sin \theta-2 \sin \theta=0$
$\alpha+\beta=-\sin \theta$
$\alpha \beta=-2 \sin \theta$
$\text { Now, } \frac{\alpha^{12}+\beta^{12}}{\left(\frac{1}{\alpha^{12}}+\frac{1}{\beta^{12}}\right)(\alpha-\beta)^{24}}$
$=\frac{(\alpha \beta)^{12}}{(\alpha-\beta)^{24}}$
$=\frac{(\alpha \beta)^{12}}{\left((\alpha+\beta)^{2}-4 \alpha \beta\right)^{12}}$
$=\left[\frac{\alpha \beta}{(\alpha+\beta)^{2}-4 \alpha \beta}\right]^{12}$
$=\left(\frac{-2 \sin \theta}{\sin ^{2} \theta+8 \sin \theta}\right)^{12}$
$=\frac{2^{12}}{(\sin \theta+8)^{12}}$
