Question : If $\cos\theta+\sin\theta=\sqrt{2}\cos\theta$, then $\cos\theta-\sin\theta$ is:
Option 1: $\sqrt{2}\tan\theta$
Option 2: $-\sqrt{2}\cos\theta$
Option 3: $-\sqrt{2}\sin\theta$
Option 4: $\sqrt{2}\sin\theta$
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Correct Answer: $\sqrt{2}\sin\theta$
Solution :
Given: $\cos\theta+\sin\theta=\sqrt{2}\cos\theta$
Squaring on both sides, we get,
$\cos^{2}\theta+\sin^{2}\theta+2\cos\theta\sin\theta=2\cos^{2}\theta$
$⇒\cos^{2}\theta-\sin^{2}\theta=2\cos\theta\sin\theta$
$⇒(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)=2\cos\theta\sin\theta$
$⇒\sqrt{2}\cos\theta(\cos\theta-\sin\theta)=2\cos\theta\sin\theta$
$\therefore \cos\theta-\sin\theta=\sqrt{2}\sin\theta$
Hence, the correct answer is $\sqrt{2}\sin\theta$.
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