A) \[4{{x}^{2}}-4x-1=0\]
B) \[4{{x}^{2}}-2x-1=0\]
C) \[8{{x}^{2}}-4x-3=0\]
D) None of these
Correct Answer: C
Solution :
\[\tan (\pi cos\theta )=cot(\pi sin\theta )\] \[\Rightarrow \]\[\pi cos\theta =n\pi +\frac{\pi }{2}-\pi \sin \theta \] \[\Rightarrow \]\[\sin \theta +\cos \theta =n+\frac{1}{2};n\in I\] \[\Rightarrow \]\[\sin \theta +\cos \theta =\frac{1}{2}or-\frac{1}{2}\] \[\Rightarrow \]\[\sin \theta \cos \theta =-\frac{3}{8}\] \[\therefore \]\[\sin \theta \]and\[\cos \theta \]are roots of \[{{x}^{2}}\pm \frac{1}{2}\times -\frac{3}{8}=0\]
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