A) \[\frac{1}{2}\]
B) \[0\]
C) \[-1\]
D) \[2\]
Correct Answer: D
Solution :
\[2-{{\cos }^{2}}\theta =3\sin \theta \cos \theta \] Dividing both sides by\[{{\cos }^{2}}\theta \], we get \[2{{\sec }^{2}}\theta -1=3\tan \theta \] \[2(1+{{\tan }^{2}}\theta )-1=3\tan \theta \] \[\Rightarrow \] \[2+2{{\tan }^{2}}\theta -1=3\tan \theta \] \[\Rightarrow \] \[2{{\tan }^{2}}\theta -3\tan \theta +1=0\] \[\Rightarrow \] \[2{{\tan }^{2}}\theta -2\tan \theta -\tan \theta +1=0\] \[\Rightarrow \] \[2\tan \theta (\tan \theta -1)-(\tan \theta -1)=0\] \[\Rightarrow \] \[(2\tan \theta -1)(\tan \theta -1)=0\] \[\Rightarrow \] \[\tan \theta =\frac{1}{2}\]and\[1\] \[\Rightarrow \] \[\cot \theta =2\]and\[1\]
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