Q51 / 160 · AP EAMCET 2023 (18 May Shift 2) · Mathematics
If $\operatorname{cosec} \theta$, and $\cot \theta$ are the roots of $\mathrm{cx}^2+\mathrm{bx}+\mathrm{a}=0$
$(b c \neq 0)$, then $b^2\left(b^2-4 a c\right)=$
- A$-2 c^4$
- B$2 c^4$
- C$-c^4$
- Correct answerD$\mathrm{c}^4$
Solution
$\mathrm{cx}^2+\mathrm{bx}+\mathrm{a}=0$
Let $\alpha=\operatorname{cosec} \theta \& \beta=\cot \theta$
$$
\alpha+\beta=-\frac{b}{c} \& \alpha \beta=\frac{a}{c}
$$
$\begin{aligned} & \text { Now, } \alpha^2-\beta^2=(\alpha+\beta)(\alpha-\beta) \\ & \Rightarrow \operatorname{cosec}^2 \theta-\cot ^2 \theta=-\frac{b}{c} \sqrt{(\alpha+\beta)^2-4 \alpha \beta} \\ & \Rightarrow 1=-\frac{b}{c} \sqrt{\frac{b^2}{c^2}-\frac{4 a}{c}} \\ & \Rightarrow 1=-\frac{b}{c} \sqrt{\frac{b^2-4 a c}{c^2}} \Rightarrow \sqrt{\frac{b^2-4 a c}{c^2}}=-\frac{c}{b} \\ & \Rightarrow \frac{b^2-4 a c}{c^2}=\frac{c^2}{b^2} \Rightarrow b^2\left(b^2-4 a c\right)=c^4\end{aligned}$
Hence, the correct answer is option: 'D'.