A) \[\pi /6\]
B) \[\pi /4\]
C) \[\pi /3\]
D) \[2\theta \]
Correct Answer: C
Solution :
[c] : The pair is \[\left( \frac{{{\cos }^{2}}\theta }{4}+{{\sin }^{2}}\theta -\frac{1}{3} \right){{x}^{2}}+\frac{2\sin \theta }{\sqrt{3}}.xy\] \[+\frac{{{\cos }^{2}}\theta }{4}.{{y}^{2}}=0\] \[a+b=\frac{{{\cos }^{2}}\theta }{4}+{{\sin }^{2}}=\theta -\frac{1}{3}+\frac{{{\cos }^{2}}\theta }{4}\] \[=\frac{{{\cos }^{2}}\theta }{2}+{{\sin }^{2}}=\theta -\frac{1}{3}=\frac{1+3{{\sin }^{2}}\theta }{6}\] \[{{h}^{2}}-ab=\frac{{{\sin }^{2}}\theta }{3}-\left( \frac{{{\cos }^{2}}\theta }{4}+{{\sin }^{2}}\theta -\frac{1}{3} \right)\frac{{{\cos }^{2}}\theta }{4}\] \[=\frac{{{(1+3si{{n}^{2}}\theta )}^{2}}}{48},\]on simplication \[\therefore \]\[\theta ={{\tan }^{-1}}\left( \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right)={{\tan }^{-1}}\left( \frac{\frac{2(1+3si{{n}^{2}}\theta )}{4\sqrt{3}}}{\frac{1+3{{\sin }^{2}}\theta }{6}} \right)\] \[={{\tan }^{-1}}\sqrt{3}=\frac{\pi }{3}.\]
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