- Home
- Standard 11
- Mathematics
If $\alpha $ and $\beta $ are the roots of the quadratic equation, $x^2 + x\, sin\,\theta -2sin\,\theta = 0$, $\theta \in \left( {0,\frac{\pi }{2}} \right)$ then $\frac{{{\alpha ^{12}} + {\beta ^{12}}}}{{\left( {{\alpha ^{ - 12}} + {\beta ^{ - 12}}} \right){{\left( {\alpha - \beta } \right)}^{24}}}}$ is equal to
$\frac{{{2^{12}}}}{{{{\left( {\sin \,\theta + 8} \right)}^{12}}}}$
$\frac{{{2^{12}}}}{{{{\left( {\sin \,\theta - 4} \right)}^{12}}}}$
$\frac{{{2^{12}}}}{{{{\left( {\sin \,\theta - 8} \right)}^{6}}}}$
$\frac{{{2^{6}}}}{{{{\left( {\sin \,\theta + 8} \right)}^{12}}}}$
Solution
$x^{2}+x \sin \theta-2 \sin \theta=0$
$\alpha+\beta=-\sin \theta$
$\alpha \beta=-2 \sin \theta$
$\text { Now, } \frac{\alpha^{12}+\beta^{12}}{\left(\frac{1}{\alpha^{12}}+\frac{1}{\beta^{12}}\right)(\alpha-\beta)^{24}}$
$=\frac{(\alpha \beta)^{12}}{(\alpha-\beta)^{24}}$
$=\frac{(\alpha \beta)^{12}}{\left((\alpha+\beta)^{2}-4 \alpha \beta\right)^{12}}$
$=\left[\frac{\alpha \beta}{(\alpha+\beta)^{2}-4 \alpha \beta}\right]^{12}$
$=\left(\frac{-2 \sin \theta}{\sin ^{2} \theta+8 \sin \theta}\right)^{12}$
$=\frac{2^{12}}{(\sin \theta+8)^{12}}$
