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# If $\cot\theta,cosec \theta$ are the roots of the equation $ax^2+bx+c=0$ and $\Delta=b^2-4ac$,then $a^4$ equals

$\begin{array}{1 1}(A)\;bc\Delta&(B)\;c\Delta\\(C)\;b^2\Delta&(D)\;b\Delta\end{array}$

$cosec\theta+\cot\theta=-\large\frac{b}{a}$
$cosec\theta\cot\theta=\large\frac{c}{a}$
$cosec^2\theta-\cot^2\theta=1$
$\Rightarrow cosec\theta-\cot\theta=-\large\frac{a}{b}$
$\Rightarrow cosec\theta=-\large\frac{1}{2}(\large\frac{b}{a}+\frac{a}{b})$
$\cot\theta=\large\frac{1}{2}(\large\frac{a}{b}-\frac{b}{a})$
Thus $-\large\frac{1}{4}(\frac{a^2}{b^2}-\frac{b^2}{a^2})=\frac{c}{a}$
$\Rightarrow a^4-b^4=-4acb^2$
$\Rightarrow a^4=b^4-4acb^2$
$\Rightarrow a^4=b^2(b^2-4ac)=b^2\Delta$
Hence (C) is the correct answer.