If $\operatorname{cosec} heta$, and $\cot heta$ are the roots of $\mathrm{cx}^2+\mathrm{bx}+\mathrm{a}=0$ $(b c
eq 0)$, then $b^2\left(b^2-4 a c ight)=$

Question

If $\operatorname{cosec} 	heta$, and $\cot 	heta$ are the roots of $\mathrm{cx}^2+\mathrm{bx}+\mathrm{a}=0$ $(b c 
eq 0)$, then $b^2\left(b^2-4 a cight)=$

Aicharya · Accepted Answer

$\mathrm{cx}^2+\mathrm{bx}+\mathrm{a}=0$ Let $\alpha=\operatorname{cosec} 	heta \& \beta=\cot 	heta$ $ \alpha+\beta=-\frac{b}{c} \& \alpha \beta=\frac{a}{c} $ $\begin{aligned} & 	ext { Now, } \alpha^2-\beta^2=(\alpha+\beta)(\alpha-\beta) \ & \Rightarrow \operatorname{cosec}^2 	heta-\cot ^2 	heta=-\frac{b}{c} \sqrt{(\alpha+\beta)^2-4 \alpha \beta} \ & \Rightarrow 1=-\frac{b}{c} \sqrt{\frac{b^2}{c^2}-\frac{4 a}{c}} \ & \Rightarrow 1=-\frac{b}{c} \sqrt{\frac{b^2-4 a c}{c^2}} \Rightarrow \sqrt{\frac{b^2-4 a c}{c^2}}=-\frac{c}{b} \ & \Rightarrow \frac{b^2-4 a c}{c^2}=\frac{c^2}{b^2} \Rightarrow b^2\left(b^2-4 a cight)=c^4\end{aligned}$

If $\operatorname{cosec} \theta$, and $\cot \theta$ are the roots of…

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