A) \[{{p}^{2}}+{{q}^{2}}-2pr=0\]
B) \[{{p}^{2}}-{{q}^{2}}+2pr=0\]
C) \[{{p}^{2}}-{{q}^{2}}-2pr=0\]
D) \[{{p}^{2}}+{{q}^{2}}+2qr=0\]
Correct Answer: B
Solution :
Since, \[\sin \alpha \] and \[\cos \alpha \] are the roots of the equation\[p{{x}^{2}}+qx+r=0\] \[\therefore \] \[\sin \alpha +\cos \alpha =-\frac{q}{p}\] ... (i) and \[\sin \alpha \cdot \cos \alpha =\frac{r}{p}\] ... (ii) On squaring Eq. (i), we get \[{{(\sin \alpha +\cos \alpha )}^{2}}={{\left( -\frac{p}{q} \right)}^{2}}\] \[\Rightarrow \]\[{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha +2\sin \alpha \cos \alpha =\frac{{{q}^{2}}}{{{p}^{2}}}\] \[\Rightarrow \] \[1+2\cdot \frac{r}{p}=\frac{{{q}^{2}}}{{{p}^{2}}}\] \[\Rightarrow \] \[\frac{(p+2r)}{p}=\frac{{{q}^{2}}}{{{p}^{2}}}\] \[\Rightarrow \] \[p(p+2r)={{q}^{2}}\] \[\Rightarrow \] \[{{p}^{2}}-{{q}^{2}}+2rp=0\]
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