MHT CET · Maths · Trigonometric Equations
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 \(\mathrm{P} \subset \mathrm{Q}\) and \(\mathrm{Q}-\mathrm{P} \neq \phi\)
- B \(\mathrm{Q} \not \subset \mathrm{P}\)
- C \(\mathrm{P} \not \subset \mathrm{Q}\)
- D \(\mathrm{P}=\mathrm{Q}\)
Answer & Solution
Correct Answer
(D) \(\mathrm{P}=\mathrm{Q}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \sin \theta-\cos \theta=\sqrt{2} \cos \theta \\ & \Rightarrow \sin \theta=(\sqrt{2}+1) \cos \theta \\ & \Rightarrow \frac{\sin \theta}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1}=\cos \theta\end{aligned}\)
\(\begin{aligned} & \quad \Rightarrow \frac{\sin \theta(\sqrt{2}-1)}{(\sqrt{2})^2-1^2}=\cos \theta \\ & \Rightarrow \frac{\sin \theta(\sqrt{2}-1)}{2-1}=\cos \theta \\ & \Rightarrow \quad \sqrt{2} \sin \theta-\sin \theta=\cos \theta \\ & \Rightarrow \\ & \Rightarrow \sin \theta+\cos \theta=\sqrt{2} \sin \theta \\ & \therefore \quad\end{aligned}\)
\(P=Q\)
\(\begin{aligned} & \quad \Rightarrow \frac{\sin \theta(\sqrt{2}-1)}{(\sqrt{2})^2-1^2}=\cos \theta \\ & \Rightarrow \frac{\sin \theta(\sqrt{2}-1)}{2-1}=\cos \theta \\ & \Rightarrow \quad \sqrt{2} \sin \theta-\sin \theta=\cos \theta \\ & \Rightarrow \\ & \Rightarrow \sin \theta+\cos \theta=\sqrt{2} \sin \theta \\ & \therefore \quad\end{aligned}\)
\(P=Q\)
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