Question : If $\operatorname{cos} \theta+\operatorname{sin} \theta=\sqrt{2} \operatorname{cos} \theta$, find the value of $(\cos \theta-\operatorname{sin} \theta)$
Option 1: $\sqrt{2} \sin \theta$
Option 2: $\sqrt{2} \cos \theta$
Option 3: $\frac{1}{\sqrt{2}} \sin \theta$
Option 4: $\frac{1}{2}\cos \theta$
Correct Answer: $\sqrt{2} \sin \theta$
Solution :
Given: $\cos \theta + \sin \theta = \sqrt{2} \cos \theta$
Squaring both sides, we get:
⇒ $\cos^2 \theta + \sin^2 \theta + 2\sin \theta \cos \theta = 2\cos^2 \theta$
⇒ $\cos^2 \theta - \sin^2 \theta = 2\cos \theta \sin \theta$
⇒$(\cos \theta + \sin \theta)(\cos \theta - \sin \theta) = 2\cos \theta \sin \theta$
Putting $\cos \theta + \sin \theta = \sqrt{2} \cos \theta$, we get:
$\sqrt{2}\cos \theta (\cos \theta - \sin \theta) = 2\cos \theta \sin \theta$
⇒ $\cos \theta - \sin \theta = \sqrt{2} \sin \theta$
Hence, the correct answer is $\sqrt{2} \sin \theta$.
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