A) \[P\subset Q\,and\,Q-P\ne \phi \]
B) \[Q\subset P\]
C) \[P\subset Q\]
D) \[P=Q\]
Correct Answer: D
Solution :
| \[P=\{\theta :\sin \theta -\cos \theta =\sqrt{2}\cos \theta \}\] |
| \[\sin \theta =\left( \sqrt{2}+1 \right)\cos \theta \Rightarrow \tan \theta =\sqrt{2}+1\] |
| \[Q=\{\theta :\sin \theta +\cos \theta =\sqrt{2}\sin \theta \}\] |
| \[\cos \theta =\left( \sqrt{2}-1 \right)\sin \theta \Rightarrow \tan \theta =\sqrt{2}+1\] |
| \[\therefore P=Q\] |
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