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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,
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Verified Answer
The correct answer is:
$P=Q$
$P=Q$
$$
\begin{aligned}
& P=\{\theta: \sin \theta-\cos \theta=\sqrt{2} \cos \theta\} \\
& \Rightarrow \quad \cos \theta(\sqrt{2}+1)=\sin \theta \\
& \Rightarrow \quad \tan \theta=\sqrt{2}+1 \\
& Q=\{\theta: \sin \theta+\cos \theta=\sqrt{2} \sin \theta\} \\
& \Rightarrow \quad \sin \theta(\sqrt{2}-1)=\cos \theta \\
& \Rightarrow \quad \tan \theta=\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} \\
& =(\sqrt{2}+1) \\
& \therefore \quad P=Q \\
&
\end{aligned}
$$
\begin{aligned}
& P=\{\theta: \sin \theta-\cos \theta=\sqrt{2} \cos \theta\} \\
& \Rightarrow \quad \cos \theta(\sqrt{2}+1)=\sin \theta \\
& \Rightarrow \quad \tan \theta=\sqrt{2}+1 \\
& Q=\{\theta: \sin \theta+\cos \theta=\sqrt{2} \sin \theta\} \\
& \Rightarrow \quad \sin \theta(\sqrt{2}-1)=\cos \theta \\
& \Rightarrow \quad \tan \theta=\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} \\
& =(\sqrt{2}+1) \\
& \therefore \quad P=Q \\
&
\end{aligned}
$$
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