If operatorname{cosec} theta and cot theta ar | vedclass

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(D) Given the quadratic equation $cx^2+bx+a=0$. Let the roots be $\alpha = \operatorname{cosec} 	heta$ and $\beta = \cot 	heta$. From the relation b

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$c^4$

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(D) Given the quadratic equation $cx^2+bx+a=0$. Let the roots be $\alpha = \operatorname{cosec} 	heta$ and $\beta = \cot 	heta$. From the relation between roots and coefficients,we have: $\alpha + \beta = -\frac{b}{c}$ and $\alpha \beta = \frac{a}{c}$. We know the identity $\operatorname{cosec}^2 	heta - \cot^2 	heta = 1$,which implies $\alpha^2 - \beta^2 = 1$. This can be written as $(\alpha + \beta)(\alpha - \beta) = 1$. Since $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta$,we have $\alpha - \beta = \pm \sqrt{(\alpha + \beta)^2 - 4\alpha \beta}$. Substituting the values: $1 = \left(-\frac{b}{c}ight) \left(\pm \sqrt{\frac{b^2}{c^2} - \frac{4a}{c}}ight)$. Squaring both sides: $1 = \frac{b^2}{c^2} \left(\frac{b^2 - 4ac}{c^2}ight)$. $1 = \frac{b^2(b^2 - 4ac)}{c^4}$. Therefore,$b^2(b^2 - 4ac) = c^4$.

If $\operatorname{cosec} \theta$ and $\cot \theta$ are the roots of $cx^2+bx+a=0$ $(bc \neq 0)$,then $b^2(b^2-4ac)=$

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