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If $\sin \alpha$ and $\cos \alpha$ are the roots of the equation $p x^{2}+q x+1$ $=0$, then which one of the following is correct?
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Verified Answer
The correct answer is:
$p^{2}-q^{2}+2 p r=0$
$\sin \alpha$ and $\cos \alpha$ are the roots of the equation $p x^{2}+q x+r=0$
$\therefore \sin \alpha+\cos \alpha=\frac{-q}{p}\cdots. (i)$
and $\sin \alpha \cos \alpha=\frac{r}{p}$ ...(ii)
From equation (i)
$(\sin \alpha+\cos \alpha)^{2}=\frac{q^{2}}{p^{2}}$
$\Rightarrow \sin ^{2} \alpha+\cos ^{2} \alpha+2 \sin \alpha \cos \alpha=\frac{q^{2}}{p^{2}}$
$\Rightarrow 1+2 \cdot \frac{r}{p}=\frac{q^{2}}{p^{2}}$
$\Rightarrow p^{2}+2 p r=q^{2}$
$p^{2}-q^{2}+2 p r=0$
$\therefore \sin \alpha+\cos \alpha=\frac{-q}{p}\cdots. (i)$
and $\sin \alpha \cos \alpha=\frac{r}{p}$ ...(ii)
From equation (i)
$(\sin \alpha+\cos \alpha)^{2}=\frac{q^{2}}{p^{2}}$
$\Rightarrow \sin ^{2} \alpha+\cos ^{2} \alpha+2 \sin \alpha \cos \alpha=\frac{q^{2}}{p^{2}}$
$\Rightarrow 1+2 \cdot \frac{r}{p}=\frac{q^{2}}{p^{2}}$
$\Rightarrow p^{2}+2 p r=q^{2}$
$p^{2}-q^{2}+2 p r=0$
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