Question

Let $P=\{\theta :\sin \theta –\cos \theta =\sqrt{2}\cos \theta \}$ and $Q=\{\theta :\sin \theta +\cos \theta =\sqrt{2}\sin \theta \}$ be two sets. Then :

• A. $P\subset Q\text{and}Q-P\ne \varphi$
• B. $P=Q$
• C. $Q\subset ̸P$
• D. $P\subset ̸Q$

Answer 1

The correct option is B $P=Q$

$P=\{\theta :\sin \theta –\cos \theta =\sqrt{2}\cos \theta \}\Rightarrow \sin \theta =(1+\sqrt{2})\cos \theta \Rightarrow \tan \theta =(1+\sqrt{2})Q=\{\theta :\sin \theta +\cos \theta =\sqrt{2}\sin \theta \}\Rightarrow \cos \theta =(\sqrt{2}-1)\sin \theta \Rightarrow \tan \theta =\frac{1}{\sqrt{2}-1}=(1+\sqrt{2})\therefore P=Q$