Class JEE Mathematics Sets, Relations, and Functions Q #1068
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4 Marks 2016 JEE Main 2016 (Online) 10th April Morning Slot MCQ SINGLE
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 $Q - P \neq \phi$
(B) $Q \nsubseteq P$
(C) $P \nsubseteq Q$
(D) $P = Q$
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Correct Answer: D
Explanation
Given, for set P:
$sin\theta - cos\theta = \sqrt{2}cos\theta$
$sin\theta = (\sqrt{2} + 1)cos\theta$
$cos\theta = (\sqrt{2} - 1)sin\theta$ ... (1)

Given, for set Q:
$sin\theta + cos\theta = \sqrt{2}sin\theta$
$cos\theta = (\sqrt{2} - 1)sin\theta$ ... (2)

From (1) and (2), P = Q

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