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If $\operatorname{cosec} \theta$, and $\cot \theta$ are the roots of $\mathrm{cx}^2+\mathrm{bx}+\mathrm{a}=0$
$(b c \neq 0)$, then $b^2\left(b^2-4 a c\right)=$
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$(b c \neq 0)$, then $b^2\left(b^2-4 a c\right)=$
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Verified Answer
The correct answer is:
$\mathrm{c}^4$
$\mathrm{cx}^2+\mathrm{bx}+\mathrm{a}=0$
Let $\alpha=\operatorname{cosec} \theta \& \beta=\cot \theta$
$$
\alpha+\beta=-\frac{b}{c} \& \alpha \beta=\frac{a}{c}
$$
$\begin{aligned} & \text { Now, } \alpha^2-\beta^2=(\alpha+\beta)(\alpha-\beta) \\ & \Rightarrow \operatorname{cosec}^2 \theta-\cot ^2 \theta=-\frac{b}{c} \sqrt{(\alpha+\beta)^2-4 \alpha \beta} \\ & \Rightarrow 1=-\frac{b}{c} \sqrt{\frac{b^2}{c^2}-\frac{4 a}{c}} \\ & \Rightarrow 1=-\frac{b}{c} \sqrt{\frac{b^2-4 a c}{c^2}} \Rightarrow \sqrt{\frac{b^2-4 a c}{c^2}}=-\frac{c}{b} \\ & \Rightarrow \frac{b^2-4 a c}{c^2}=\frac{c^2}{b^2} \Rightarrow b^2\left(b^2-4 a c\right)=c^4\end{aligned}$
Let $\alpha=\operatorname{cosec} \theta \& \beta=\cot \theta$
$$
\alpha+\beta=-\frac{b}{c} \& \alpha \beta=\frac{a}{c}
$$
$\begin{aligned} & \text { Now, } \alpha^2-\beta^2=(\alpha+\beta)(\alpha-\beta) \\ & \Rightarrow \operatorname{cosec}^2 \theta-\cot ^2 \theta=-\frac{b}{c} \sqrt{(\alpha+\beta)^2-4 \alpha \beta} \\ & \Rightarrow 1=-\frac{b}{c} \sqrt{\frac{b^2}{c^2}-\frac{4 a}{c}} \\ & \Rightarrow 1=-\frac{b}{c} \sqrt{\frac{b^2-4 a c}{c^2}} \Rightarrow \sqrt{\frac{b^2-4 a c}{c^2}}=-\frac{c}{b} \\ & \Rightarrow \frac{b^2-4 a c}{c^2}=\frac{c^2}{b^2} \Rightarrow b^2\left(b^2-4 a c\right)=c^4\end{aligned}$
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