Trigonometrical Equations 1 Question 12
12. If $P={\theta: \sin \theta-\cos \theta=\sqrt{2} \cos \theta}$ and $Q={\theta: \sin \theta+\cos \theta=\sqrt{2} \sin \theta}$ be two sets. Then, (2011)
(a) $P \subset Q$ and $Q-P \neq \varphi$
(b) $Q \notin P$
(c) $P \not \subset Q$
(d) $P=Q$
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Answer:
Correct Answer: 12. (d)
Solution:
- $P={\theta: \sin \theta-\cos \theta=\sqrt{2} \cos \theta}$
$\Rightarrow \cos \theta(\sqrt{2}+1)=\sin \theta$
$\Rightarrow \tan \theta=\sqrt{2}+1$
$\Rightarrow Q={\theta: \sin \theta+\cos \theta}=\sqrt{2} \sin \theta$
$\Rightarrow \sin \theta(\sqrt{2}-1)=\cos \theta$
$\Rightarrow \tan \theta=\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}=(\sqrt{2}+1)$
$\therefore \quad P=Q$
