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

Answer 1

The correct option is D $P=Q$

$P:\{\theta :sin\theta -cos\theta =\sqrt{2}cos\theta \}$

square Both sides, $(sin\theta -cos\theta {)}^2-(\sqrt{2}cos\theta {)}^2$

$\Rightarrow si{n}^2\theta +co{s}^2\theta -2sin\theta cos\theta =2co{s}^2\theta$

$\Rightarrow 1-2sin\theta cos\theta =2co{s}^2\theta$

$P\Rightarrow si{n}^2\theta =co{s}^2\theta +2sin\theta cos\theta$

For Q : $\{\theta :sin\theta +cos\theta =\sqrt{2}sin\theta \}$

square $(sin\theta +cos\theta {)}^2=(\sqrt{2}sin\theta {)}^2$

$\Rightarrow si{n}^2\theta +co{s}^2\theta +2sin\theta cos\theta =2si{n}^2\theta$

$\Rightarrow 1+2sin\theta cos\theta =2si{n}^2\theta$

$Q\Rightarrow co{s}^2\theta +2sin\theta cos\theta =si{n}^2\theta$

Hence , $P=Q$