Question : If $\cos \theta-\sin \theta=\sqrt{2} \sin \theta$, then $(\cos \theta+\sin \theta)$ is:
Option 1: $-\sqrt{2} \cos \theta$
Option 2: $\sqrt{2} \cos \theta$
Option 3: $\sqrt{2} \tan \theta$
Option 4: $-\sqrt{2} \sin \theta$
Correct Answer: $\sqrt{2} \cos \theta$
Solution :
According to the question,
⇒ $\cos \theta-\sin \theta=\sqrt{2} \sin \theta$
⇒ $(\cos\theta + \sin\theta)^{2} = 2 \sin^{2}\theta$
⇒ $\cos^{2}\theta + \sin^{2}\theta + 2\sin\theta\cos~\theta = 2\sin^{2}\theta$
⇒ $\sin^{2}\theta - \cos^{2}\theta - 2\sin~\theta~cos\theta = 0$
⇒ $\sin^{2}\theta + \cos^{2}\theta - 2\sin\theta\cos\theta = 2 \cos^{2}\theta$
⇒ $(\sin~\theta - \cos\theta)^{2}=2\cos^2\theta$
⇒ $\sin\theta - \cos \theta$ = $\sqrt{2}\cos\theta$
Hence, the correct answer is $\sqrt{2}\cos\theta$.
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