A) \[{{x}^{2}}-14x+1\]
B) \[{{x}^{2}}-14x+1=0\]
C) \[{{x}^{2}}-10x+1\]
D) None of these
Correct Answer: A
Solution :
[a] \[{{\cos }^{2}}\theta +b+{{\sin }^{2}}\theta +b=-4\] \[\therefore \,\,\,\,\,\,b=-\frac{5}{2}\] \[({{\cos }^{2}}\theta +b)\,({{\sin }^{2}}\theta +b)=\frac{61}{16}\] \[\therefore \,\,\,\,\,\,{{\cos }^{2}}\theta {{\sin }^{2}}\theta +b+{{b}^{2}}=\frac{61}{16}\] \[\therefore \,\,\,\,\,\,{{\cos }^{2}}\theta {{\sin }^{2}}\theta =\frac{61}{16}+\frac{5}{2}-\frac{25}{4}=\frac{61}{16}-\frac{15}{4}=\frac{1}{16}\] \[\therefore \,\,\,\,\,\,{{\sin }^{2}}2\theta =\frac{1}{4}\] Sum of roots \[={{\tan }^{2}}\theta +{{\cot }^{2}}\theta =\frac{1-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta }{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }\] \[=\frac{1-\frac{{{\sin }^{2}}2\theta }{2}}{\frac{{{\sin }^{2}}2\theta }{4}}=\frac{1-\frac{1}{8}}{\frac{1}{16}}=14\] Product of roots = 1
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