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:

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

$\Rightarrow \cos \theta(\sqrt{2}+1)=\sin \theta$

$\Rightarrow \tan \theta=\sqrt{2}+1$

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

$\Rightarrow \sin \theta(\sqrt{2}-1)=\cos \theta$

$\Rightarrow \tan \theta=\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}=(\sqrt{2}+1)$

$\therefore \quad P=Q$



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