If \alpha and \beta are the roots of the quadratic equation,

x^{2}+xsin\theta -2sin\theta =0,\theta \epsilon (0,\frac{\pi}{2}), then 

\frac{\alpha ^{12}+\beta ^{12}}{(\alpha ^{-12}+\beta ^{-12})\cdot (\alpha -\beta )^{24}} is equal to :

  • Option 1)

    \frac{2^{12}}{(sin\theta-4)^{12}}

  • Option 2)

    \frac{2^{12}}{(sin\theta+8)^{12}}

  • Option 3)

    \frac{2^{12}}{(sin\theta-8)^{6}}

  • Option 4)

    \frac{2^{6}}{(sin\theta+8)^{12}}

 

Answers (1)

\alpha and \beta are the roots of the equation 

x^{2}+xsin\theta -2sin\theta =0

\alpha +\beta =-sin\theta

\alpha \beta =-2sin\theta

Now,

\frac{\alpha ^{12}+\beta ^{12}}{(\alpha ^{-12}+\beta ^{-12})\cdot (\alpha -\beta )^{24}}=\frac{(\alpha\beta)^{12}}{(\alpha-\beta)^{24}}

                                                   =\frac{(\alpha \beta )^{12}}{[(\alpha +\beta )^{2}-4\alpha \beta ]^{12}}

                                                      =[\frac{-2sin\theta}{sin^{2}\theta+8sin\theta}]^{12}

                                                      =\frac{2^{12}}{(sin\theta+8)^{12}}

correct option (2)


Option 1)

\frac{2^{12}}{(sin\theta-4)^{12}}

Option 2)

\frac{2^{12}}{(sin\theta+8)^{12}}

Option 3)

\frac{2^{12}}{(sin\theta-8)^{6}}

Option 4)

\frac{2^{6}}{(sin\theta+8)^{12}}

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