A) \[{{a}^{2}}-{{b}^{2}}+2ac=0\]
B) \[{{a}^{2}}+{{b}^{2}}+2ac=0\]
C) \[\,{{a}^{2}}-{{b}^{2}}-2ac=0\]
D) \[{{a}^{2}}+{{b}^{2}}-2ac=0\]
Correct Answer: A
Solution :
Given \[\sin \theta ,\,\cos \,\theta \] are the roots of the equation \[a{{x}^{2}}+bx+c=0\,\,\,a\ne 0.\] Then, sum of the roots \[=\sin \theta +\cos \theta =-b/a\] ?.(i) Product of the roots \[=\sin \theta .\cos \theta =c/a\] ?.(ii) From Eq. (i), \[{{(\sin \theta +\cos \theta )}^{2}}={{b}^{2}}/{{a}^{2}}\] \[\Rightarrow \]\[({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )+2\sin \theta .\cos \theta ={{b}^{2}}/{{a}^{2}}\] \[\Rightarrow \] \[1+2.\,\,c/a={{b}^{2}}/{{a}^{2}}\] (from Eq. (ii)) \[\Rightarrow \] \[{{a}^{2}}+2ac={{b}^{2}}\] \[\Rightarrow \]\[{{a}^{2}}-{{b}^{2}}+2ac=0\]
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