• # question_answer 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 : [JEE Main 10-4-2019 Morning] A) $\frac{{{2}^{6}}}{{{\left( \sin \theta +8 \right)}^{12}}}$                     B) $\frac{{{2}^{12}}}{{{\left( \sin \theta -8 \right)}^{6}}}$C) $\frac{{{2}^{12}}}{{{\left( \sin \theta -4 \right)}^{12}}}$                     D) $\frac{{{2}^{12}}}{{{\left( \sin \theta +8 \right)}^{12}}}$

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