A) \[~16{{x}^{2}}-12x+1=0\]
B) \[16{{x}^{2}}+12x+1=0\]
C) \[16{{x}^{2}}-12x-1=0\]
D) \[16{{x}^{2}}+\text{10}x+1=0\]
Correct Answer: A
Solution :
Since, \[{{\sin }^{2}}{{18}^{o}}\]and \[{{\cos }^{2}}{{36}^{o}}\]are the roots of a quadratic equation. \[\therefore \]Sum of roots \[={{\sin }^{2}}{{18}^{o}}+{{\cos }^{2}}{{36}^{o}}\] \[=\left( \frac{\sqrt{5}-1}{4} \right)+{{\left( \frac{\sqrt{5}+1}{4} \right)}^{2}}\] \[=\frac{5+1-2\sqrt{5}}{16}+\frac{5+1+2\sqrt{5}}{16}\] \[=\frac{12}{16}=\frac{3}{4}\] and product of roots \[={{\sin }^{2}}{{18}^{o}}.{{\cos }^{2}}{{36}^{o}}\] \[={{\left( \frac{\sqrt{5}-1}{4} \right)}^{2}}{{\left( \frac{\sqrt{5}+1}{4} \right)}^{2}}\] \[={{\left( \frac{5-1}{4\times 4} \right)}^{2}}={{\left( \frac{1}{4} \right)}^{2}}=\frac{1}{16}\] \[\therefore \]Required equation whose roots are \[{{\sin }^{2}}{{18}^{o}}\]and \[{{\cos }^{2}}{{36}^{o}},\]is \[{{x}^{2}}-\](sum of roots) \[x+\](product of roots) \[=0\] \[\Rightarrow \]\[{{x}^{2}}-\frac{3}{4}x+\frac{1}{16}=0\] \[\Rightarrow \]\[16{{x}^{2}}-12x+1=0\]
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